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SH6610 instructions introduction SH6610 instructions introduction sino w ealth electronic (shanghai) l t d. ve r 1 . 0 1 / 3 6 http://
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n analysis of SH6610 instructions the following a r e SH6610 in struct ions, catego rized a nd ex plaine d acco rdin g to their re spe c tive function s. wh en you nee d an instruct i o n for a certai n function, you can loo k up the instru ctio n in its function categ o ry. you?d better browse throu gh all the instru ction s listed here, beca u s e even thoug h you can?t rememb er all of them at once, you can still have an impre s sion to remind you of such inst ruct ion when need ed. some people don ?t even know about som e very good inst ruction s that he can us e, just beca u se he didn?t browse throu gh all of them. this igno ra nce may cau s e consi d e r abl e wa ste of spa c e on a syste m without a b i g enou gh rom , which is so reg r ettable. i will ex plain as simply a s possibl e to i m prove yo u r learning effici ency. of co urse, it is n o r mal to u nde rsta nd o r re membe r o n ly part of the conte n ts afte r re adin g on ce. and you are ce rtain of ten reve rt to this bo ok be cau s e it i s a colle ct ion of i n st r u ct io ns. instructions instru ction i s a se rie s of cod e s that can be re cog n ize d by cp u, and th en cpu will ope rate a c cording to the gi ven instructio n. opera n d ap art from i n structio ns t o tell cpu how to op erate, the ope ration o b je ct must be desi gnate d . the o b je ct for cp u to o p e rate with i s calle d opera n d . therefo r e, a co mplet e instructio n must incl ude two items, both inst r u ct io n and ope ran d . form at of SH6610 in stru cti ons instru ction [operand 1], [operand 2] of the above, items insi de [ ] are used acco rdin g to nature of the instructio n. some instructio ns requi re only o ne ope ra nd, while som e requi re two. in stru ction a nd ope ran d sh all be se p a rate d by a sp a c e, a nd ope ra nd s shall b e se p a rated fro m ea ch othe r by a ?,?. execution t i me of instru ction executio n time for SH6610 in stru cti ons i s one in stru ction cy cl e , which is on e fourth of the system worki ng freq uen cy . ve r 1 . 0 2 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n 1-1 instruction categories accordi ng to their fun c tion s, SH6610 in st ru ction s ca n be cla s sified i n to four categ o rie s : z arithmetic o p eratio n instru ction s i. addition: adc d adcm d ad d d addm d adi d adi m ii. subtra ction: sbc d sbcm d sub d su b m d sbi d sbim iii. bcd: daa d das z logi c ope r ati on instructio n s eor d eorm d eorim d or d orm d orim d an d d andm d a ndim z dat a t r a n smi ssi on i n st ru ct ions lda d st a d ldi z flow control instru ction s baz d bc d ba0 d ba1 d ba2 d ba3 d call d rt n w d rt n i d hal t d st op d jmp d tjmp the above a r e all of the instructio ns o f SH6610 se rie s , which a dd up to onl y 40 in numb e r , b u t never overl o ok the m ! a variety of co nsu m er ele c t r ical prod uct s on the market are cre a ted with them, e.g. cal c ulato r , remote c o ntroller , wat c h, toy , etc . 1-2 explanation of sy mbols before goin g to our su bje c t, we li st the sym bols th at may app ea r af terwards so that our rea ders can u nde rst and thi s boo k mo re easily . th e sy mbol s are li sted as follo ws: pc program counter ac ac cumulator cy ca rry flag mx dat a m e mory bbb ram ban k st s t ack tbr t able bran ch regi ste r x program addres s i immediate dat a ve r 1 . 0 3 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n & logi c and | logi c or ^ logi c eor no w let? s ent er the world o f instru ction s of sh661 0 se rie s . ?let ?s g o ? f ve r 1 . 0 4 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n 1-3 instructions for d a ta transmission during th e internal o p e ratio n of the system, da ta is transmit t ed rapidly a nd incessa n tly betwe en me morie s o r regi sters. this fast and in ce ssant tra nsmi s sion i s the po we r of syste m capability. SH6610 sy stem provides several in structions for data tr ansmission, as follows: ins t ruc t ion: ldi fun c tion: to load immedi ate dat a i to ?accumul ator? a nd ?dat a memory? format: instru ction code: ca rry flag: ope r ation: ldi mx d i 01 1 1 1 iiii xxx xxxx not af fec t ed ac , mx ? i explanation ldi is a very freque ntly used inst ru ction. it loads immediate d a t a i to accu mulator a n d dat a m e mo ry . ho weve r , du e to this 4-bit system , the pre s et rang e of the imme di ate dat a i is 00h ~ 0f h(0 ~ 15 ), and th at of mx is 00h~7f h . [ example ] l d i 2 0 h d 05 h a f t e r e x e c u t i o n : a c = 0 5 h content of dat a memo ry $20 h=05 h progr amming tip there isn ? t any sp eci a ll y define d re gi ster for u s e r s in sh66 10 system , but we can u s e it s powerful d a t a m e m o ry a s registe r s in ou r pro g ram designi ng. in the abo ve e x a m ple, i have use d dat a m e m o ry $20 h fo r a regi ste r . ho we ve r , if th ey a r e e x pressed o n ly b y a ddress an d without respe c tive n a m e s, the desi gn of the pro g ram will be very confusi ng. here ? s a tip for you: you can use the p s eu do-i n structio n equ to defin e each dat a a ddress. a a a e q u 2 0 h n a m e o f m e mo ry v a ri a b l e dat a memor y address pseud o- therefore, the ldi inst ru ction ca n also be written lik e this : l d i a a a d 05h ve r 1 . 0 5 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: s t a fun c tion:to st ore the valu e of accu mulat o r to dat a me mory format: instru ction code: ca rry flag: ope r ation: st a mx d bbb 001 1 1 1bbb x xx xxxx not af fec t ed mx ? ac explanation st a loads the val ue o f accumul a tor to dat a m e mory . whe n executin g thi s in stru ction, cpu d o e s the tran smissio n only and th e ca rry flag is not af fected. [ example ] save the value of ac in $ 21h : l d i 2 0 h d 05 h ;ac=0 5h d $ 20h=0 5h s t a 2 1 h d 00 h ;$21h=0 5h : progr amming tip if operan d 2 is the immediate dat a i, then the prog ram can be written in the followin g wa ys: l d i 2 0 h , 0ah ; e x presse d in hex l d i 2 0 h , 10 ; e x presse d in decim al l d i 2 0 h , 1010 b ; e x presse d in binary i n st ru ct ion: l d a fun c tion: to load the value of dat a mem o ry to accum u lator format: instru ction code: ca rry flag: ope r ation: lda mx d bbb 001 1 1 0bbb x xx xxxx not af fec t ed ac ? mx explanation lda loa d s th e value of da t a memo ry to accumul a tor . this instru ction do es no t af fect the ca rry flag. ve r 1 . 0 6 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n [ example ] load the valu e of $20h to ac : ldi 20h,0 5h ;$20h=0 5h,a c=05 h ldi 21h,0 fh ;$21h=0 fh,a c=0f h lda 20h,0 ;ac=0 5h : progr amming tip when writi ng a prog ra m , we often use la bel s a s jum p ing de stination s in the pro g ram . label n a m e s can b e define d by u s e r accordi ng to the followi ng rule s: i. a label m u stn ? t begin with num ber or sp a c e. ii. length of a label m u stn ? t exce ed 7 characte rs. only the first 7 chara c ters will be re cog n ized fo r label s e x ce eding that len g th. ve r 1 . 0 7 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n 1-4 instructions for a r ithmetic operation sh66 10 se rie s provid e so me freque ntly used ar ithm e t ic (intege r) o peration instruction s , like: addition, subtractio n, bcd a d ju stment, etc. i n st ru ct ion: a dd fun c tion: to add up the val ues of dat a memo ry and accumulato r , and then sav e the re sult in the accu mul a tor format: instru ction code: ca rry flag: ope r ation: add mx d bbb 00001 0bbb xxx xxxx cy ac ? mx + a c explanation instru ctio n add a d d s up the value s of dat a mem o ry and a c cu mulator a nd saves the re sult in the a c cumul a tor . the add op e r ati on af fe ct s ca rry flag: wh en the re sult of add exc e eds 0fh, the ca rry flag is s e t to 1; otherwi se the value of cy is 0. therefo r e we ca n deci de wheth e r there? s a carry by the va lue of the ca rry flag af ter ad dition. [ example ] 05h +06 h : ldi 20h,0 5h ;$20h=0 5h,a c=05 h ldi 21h,0 6h ;$21h=0 6h,a c=06 h lda 20h,0 0h ;ac=0 5h add 21h,0 ;ac=0b h,cy =0,$2 1h = 0 6 h : [ example ] 0bh + 06h : ldi 20h,0b h ;$20h=0b h , a c=0bh ldi 21h,0 6h ;$21h=0 6h,a c=06 h lda 20h,0 0h ;ac=0b h,cy =0 a d d 2 1 h , 0 ; a c = 0 1 h , c y = 1 , $ 2 1 h = 0 6 h : i n st ru ct ion: a d d m fun c tion: to add up the val ues of dat a memo ry and accumulato r , and save the re sult in both the accumul a tor and the dat a memory format: instru ction code: ca rry flag: ope r ation: addm mx d bbb 00001 1bbb xxx xxxx cy ac , mx ? mx + ac ve r 1 . 0 8 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n explanation instru ctio n addm a d d s up the valu es of dat a m e mory a nd a c cumul a tor a nd saves the re sult in b o th the accu mula tor an d the d a t a mem o ry . the addm ope ration af fect s ca rry flag: whe n the re sult of addm excee d s 0f h, the ca rry fla g is set to 1; otherwi se the value cy i s 0. therefore we can de cid e whethe r there? s a carry by the value of the carry flag af ter ad dition. [ example ] 05h +06 h : ldi 20h,0 5h ;$20h=0 5h,a c=05 h ldi 21h,0 6h ;$21h=0 6h,a c=06 h l d a 2 0 h , 0 0 h ; a c = 0 5 h addm 21h,0 ;ac=0b h,$2 1h=0b h ,cy=0 : [ example ] 0bh + 06h : ldi 20h,0b h ;$20h=0b h , a c=0bh ldi 21h,0 6h ;$21h=0 6h,a c=06 h l d a 2 0 h , 0 0 h ; a c = 0 b h , c y = 0 a d d m 2 1 h , 0 ; a c = 0 1 h , $ 2 1 h = 0 1 h , c y = 1 : progr amming tip when yo u are readi ng the exam ples, i sugge st tha t you call the m into ice af ter com p iling to watch the cha nge in e a c h of the re gi sters ste p by step. i n st ru ct ion: a dc fun c tion: to add up the val ue of dat a m e mory , carry flay and the value of accu mulato r , and then save the re sult in the ac cumulator format: instru ction code: ca rry flag: ope r ation: adc mx d bbb 00000 0bbb xxx xxxx cy ac ? mx + a c + cy explanation instru ctio n adc ad ds up the value of dat a memo ry , carry flag and the value of ve r 1 . 0 9 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n accum u lato r , and saves th e re sult in the accu m u lato r . the adc ope ration af f e ct s the carry flag: when th e re sult of adc exce ed s 0 fh, the ca rry flag is set to 1; otherwise the value of c y is 0. theref ore we can d e cid e wheth e r there? s a carry by the value of the carry flag af ter addition. [ example ] 05h +06 h , cy=1 : ; c y = 1 ldi 20h,0 5h ;$20h=0 5h,a c=05 h ldi 21h,0 6h ;$21h=0 6h,a c=06 h l d a 2 0 h , 0 0 h ; a c = 0 5 h adc 21h,0 ;ac=0 c h,$2 1h=06 h ,cy = 0 : [ example ] 0bh + 06h , cy=0 : ; c y = 0 ldi 20h,0b h ;$20h=0b h , a c=0bh ldi 21h,0 6h ;$21h=0 6h,a c=06 h l d a 2 0 h , 0 0 h ; a c = 0 b h a d c 2 1 h , 0 ; a c = 0 1 h , $ 2 1 h = 0 6 h , c y = 1 : progr amming tip when carry i s n o t co n s ide r e d , yo u ? d better use add rathe r t han a dc, in ord e r to a v oi d extra un cert ai nty due to the addition of cy value (be c a u se cy ca n b e either 1 o r 0). v e r1 .0 10/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: a d c m fun c tion: to add up the val ue of dat a m e mory , carry flag an d the value of accu mulato r , and then save the re sult in both the accumulato r and the dat a memo ry format: instru ction code: ca rry flag: ope r ation: adcm mx d bbb 00000 1bbb xxx xxxx cy ac , mx ? mx + a c + cy explanation instru ctio n adcm ad ds up the value of dat a memory , ca rry flag and the value of accum u lato r , and saves th e re sult in bot h the accum u lator an d the dat a mem o ry . the adcm ope ration af fect s the ca rry flag: wh en t he result of adcm exceed s 0f h, the carry flag is set to1; otherwi se the value of cy is 0. th eref o r e we ca n de cide wh e t her the r e? s a ca rry by the value of the carry flag af te r addition. [ example ] 05h +06 h , cy=1 : ; c y = 1 ldi 20h,0 5h ;$20h=0 5h,a c=05 h ldi 21h,0 6h ;$21h=0 6h,a c=06 h l d a 2 0 h , 0 0 h ; a c = 0 5 h adcm 21h,0 ;ac=0 c h,$2 1h =0 ch, cy= 0 : [ example ] 0bh + 06h , cy=0 : ; c y = 0 ldi 20h,0b h ;$20h=0b h , a c=0bh ldi 21h,0 6h ;$21h=0 6h,a c=06 h lda 20h,0 0h ;ac=0b h, a d c m 2 1 h , 0 ; a c = 0 1 h , $ 2 1 h = 0 1 h , c y = 1 : i n st ru ct ion: a d i fun c tion: to add up the val ue of dat a m e mory an d immediate dat a i, and then sa ve the result in accumul a to r format: instru ction code: ca rry flag: ope r ation: adi mx d i 01000 iiii xxx xxxx cy ac ? mx + i explanation instru ctio n adi adds u p the value of dat a memory and immedi ate dat a i, and save s the ve r 1 . 0 1 1 / 3 6
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n re sult in accu mulato r . the adi operati on af fect s carry flag: when the result of adi exceed s 0fh, the carry flag is set to1; otherwi se the va lu e of cy is 0. therefo r e we ca n de cide wheth e r the r e ? s a ca rry by the val ue of the ca rry flag af ter additio n . [ example ] $20h=0 5h , i=04 h : ldi 20h,0 5h ;$20h=0 5h,a c=05 h,cy =0 a d i 2 0 h , 0 4 h ; a c = 0 9 h , $ 2 0 h = 0 5 h , c y = 0 : [ example ] $20h=0 ah , i=07h : ldi 20h,0a h ;$20h=0a h , a c=0ah,cy =0 a d i 2 0 h , 0 7 h ; a c = 0 1 h , $ 2 0 h = 0 a h , c y = 1 : i n st ru ct ion: a d i m fun c tion: to add up the val ue of dat a m e mory an d immediate dat a i, and then sa ve the result in both accum u lator a nd the dat a m e mo ry format: instru ction code: ca rry flag: ope r ation: a d i m m x d i 01001 iiii xxx xxxx cy ac , mx ? mx + i explanation instruction adim add s up the value of dat a memory and immediate dat a i, and saves th e re sult in both accumulato r and the dat a memory . the adim o peration af fect s carry flag: whe n the result of adim e x ceed s 0 f h, the ca rry fl ag is set to1; oth e rwise the val ue of cy i s 0. therefore we can de cid e b y the value of the ca rry flag af ter additio n whethe r there is a ca rry . [ example ] $20h=0 5h , i=04 h : ldi 20h,0 5h ;$20h=0 5h,a c=05 h a d i m 2 0 h , 0 4 h ; a c = 0 9 h , $ 2 0 h = 0 9 h , c y = 0 : [ example ] $20h=0 ah , i=07h : ldi 20h,0a h ;$20h=0a h , a c=0ah v e r1 .0 12/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n a d i m 2 0 h , 0 7 h ; a c = 0 1 h , $ 2 0 h = 0 1 h , c y = 1 : i n st ru ct ion: d a a func tion: to adjust the valu e of dat a me mory to deci m al af ter addition, an d then save the result in both accu mulato r and the dat a memo ry format: instru ction code: ca rry flag: ope r ation: daa mx 1 1 001 01 10 x xx x xxx cy ac ; mx ? dec i mal , adjust ac for add explanation instru ctio n daa act s b y adjusting th e value of dat a memory to deci m al af ter addition an d saving the re sult to both accumul a tor a nd the dat a memo ry . it s adjusting m e thod is if the value of the dat a m e mo ry is g r eate r th an 9 or if cy = 1, the n ad d 6 to the dat a memo ry and s e t the c a rry flag to 1. [ example ] 06h + 05 h , and do daa adjustme n t : l d i 2 0 h , 0 6 h ; a c = 0 6 h , $ 2 0 h = 0 6 h l d i 2 1 h , 0 5 h ; a c = 0 5 h , $ 2 1 h = 0 5 h l d a 2 0 h , 0 ; a c = 0 6 h a d d 2 1 h , 0 ; a c = 0 b h , c y = 0 d a a 2 1 h ; a c = 0 1 h , $ 2 1 h = 0 1 h , c y = 1 : v e r1 .0 13/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: s u b fun c tion: to subtra ct the value of accum u lator fro m the value of dat a m e mo ry , and then sa ve the result in the ac cumulator format: instru ction code: ca rry flag: ope r ation: sub mx d bbb 0001 1 0bbb xxx x xxx cy ac ? mx - ac explanation sub subtract s the value of accumulato r from the valu e of dat a me mory and sav e s the re sult in the a c cumul a tor . whe n exe c uti ng sub, if the value of d a t a memo ry is l e ss than th e value of accumulator , a ?borrow? will t a ke pl ace and cy will be set to 0. on the contrary , i f the value of dat a mem o ry is gre a ter th an the value of accumul a tor , bo rro w wil l not happe n and cy will be set to 1. therefo r e we ca n d e ci de by the value of cy wheth e r the r e i s a borro w af te r executio n of sub. besid e s, subtractio n in the syste m is do ne th rough additio n , i.e. when sub t ractin g a nu mbe r , it is actually addin g the num ber ? s bina ry compl e ment . [ example ] 06h - 05h : ldi 20h,0 5h ;ac=0 5 h,$ 2 0 h =05 h ldi 21h,0 6h ;ac=0 6 h,$ 2 1 h =06 h l d a 2 0 h , 0 ; a c = 0 5 h sub 21h,0 ;ac=1, cy=1, $21 h=06 h : [ example ] 05h - 06h : ldi 20h,0 5h ;ac=0 5 h,$ 2 0 h =05 h ldi 21h,0 6h ;ac=0 6 h,$ 2 1 h =06 h l d a 2 1 h , 0 ; a c = 0 6 h s u b 2 0 h , 0 ; a c = 0 f h , c y = 0 , $ 2 0 h = 0 5 h : v e r1 .0 14/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: s u b m fun c tion: to subtra ct the value of accum u lator fro m the value of dat a m e mo ry , and then sa ve the result in both the accumulato r and the dat a memo ry format: instru ction code: ca rry flag: ope r ation: subm mx d bbb 0001 1 1bbb xxx x xxx cy ac , mx ? mx - ac explanation system movement of subm is alm o st the sam e as sub, it subt ra ct s current value of accum u lator f r om the val u e of dat a me mory an d saves the result in the accumulator as well as in the dat a memory . when exe c utin g subm, if the value of dat a memo ry is l e ss than the value of accumulator , a ?borrow? will t a ke pl ace and cy will be set to 0. on the contrary , i f the value of dat a mem o ry is gre a ter th an the value of accumul a tor , bo rro w wil l not happe n and cy will be 1. there f ore we ca n deci de by the value of cy whethe r there i s a borr o w af ter exe c utio n of subm. [ example ] 06h - 05 h : ldi 20h,0 5h ;ac=0 5 h,$ 2 0 h =05 h ldi 21h,0 6h ;ac=0 6 h,$ 2 1 h =06 h lda 20h,0 ;ac=0 5h su bm 2 1 h ,0 ;ac = 01 h , c y = 1 ,$ 21 h= 01h : [ example ] 05h - 06 h : ldi 20h,0 5h ;ac=0 5 h,$ 2 0 h =05 h ldi 21h,0 6h ;ac=0 6 h,$ 2 1 h =06 h l d a 2 1 h , 0 ; a c = 0 6 h subm 20h,0 ;ac=0f h,cy =0,$2 0 h = 0 f h : v e r1 .0 15/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: s b c fun c tion: to subtra ct the value of accum u lator fro m the value of dat a m e mo ry , add ca rry fl ag, and then save the resu lt in the accumul a tor format: instru ction code: ca rry flag: ope r ation: sbc mx d bbb 00010 0bbb xxx xxxx cy ac ? mx - ac + cy explanation system mov e ment of sb c is to su btract the valu e of accum u la tor from th e value of dat a mem o ry , add the valu e of carry flag, and then sa ve the result i n the accumu lator . whe n executin g sbc, if the value of dat a memory is le ss than the value of accum u lat o r , a ?bo r row? will t a ke pla c e and the cy will be set to 0. on the cont rary , if th e value of dat a memory is gre a ter tha n the value of accum u lat o r , bo rrow will not hap pe n and th e cy will be 1. therefore we can d e ci de by the value of cy wheth e r there is a borrow af ter execution o f sbc. [ example ] cy=0 d 6 - 5 = ? : ; c y = 0 ldi 20h,0 5h ;ac=0 5 h,$ 2 0 h =05 h ldi 21h,0 6h ;ac=0 6 h,$ 2 1 h =06 h lda 20h,0 ;ac=0 5 h sbc 2 1 h ,0 ;ac = 01 h , c y = 0 ,$ 21 h= 06h : [ example ] cy=1 d 6 - 5=? : ;cy= 1 ldi 20h,0 5h ;ac=0 5 h,$ 2 0 h =05 h ldi 21h,0 6h ;ac=0 6 h,$ 2 1 h =06 h lda 20h,0 ;ac=0 5h sbc 2 1 h ,0 ;ac = 02 h , c y = 0 ,$ 21 h= 06h : i n st ru ct ion: s b c m fun c tion: to subtra ct the value of accum u lator fro m the value of dat a m e mo ry , add ca rry fl ag, and then save the resu lt in both the accu mulator a nd the dat a mem o ry format: instru ction code: ca rry flag: ope r ation: sbcm mx d bbb 00010 1bbb xxx xxxx cy ac , mx ? mx - ac + cy v e r1 .0 16/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n explanation system movement of sbcm is to su btra ct the valu e of accum u l a tor from th e value o f dat a me mory , add the valu e of ca rry fla g , and the n save the re sul t in both the accum u lato r and dat a m e mory . when executin g s b cm, if the value of d a t a memo ry is l e ss than the value of accu mulato r , a ?b orrow? will t a ke pla c e and the cy will b e set to 0. on the contra ry , if the value of dat a m e mo ry is greater t han the val u e of accum u lat o r , bo rrow wil l not hap pen and the cy will be set to 1. therefo r e we can de cid e by the valu e of cy whet her th ere i s a borro w af ter executio n of sbcm. [ example ] cy=0 d 6 - 5=? : ; c y = 0 ldi 20h,0 5h ;ac=0 5 h,$ 2 0 h =05 h ldi 21h,0 6h ;ac=0 6 h,$ 2 1 h =06 h l d a 2 0 h , 0 ; a c = 0 5 h , c y = 0 sbc m 2 1 h ,0 ;ac = 01 h , c y = 0 ,$ 21 h= 01h : [ example ] cy=1 d 6 - 5=? : ;cy= 1 ldi 20h,0 5h ;ac=0 5 h,$ 2 0 h =05 h ldi 21h,0 6h ;ac=0 6 h,$ 2 1 h =06 h l d a 2 0 h , 0 ; a c = 0 5 h , c y = 1 s b c m 2 1 h , 0 ; a c = 0 2 h , c y = 0 , $ 2 1 h = 0 2 h : i n st ru ct ion: s b i fun c tion: to subtra ct the immediat e dat a i from the value of dat a memo ry , and then save the result in accumulato r format: instru ction code: ca rry flag: ope r ation: sbi mx d i 01010 iiii xxx xxxx cy ac ? mx - i explanation system movemen t of sbi is to subtract i mmediate d a t a i from the value of dat a memo ry , and save the re sult in accu mulato r . wh en exec uting sbi, if the value of the dat a memo ry is less than the immediate d a t a , a ?bo rro w? will t a ke place and cy will be set to 0. on th e v e r1 .0 17/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n cont rary , if the value of dat a memory is gr eater than the immediate dat a, borrow will not happ en an d cy will be 1. therefo r e we ca n de ci d e by the valu e of cy whether there is a borro w af ter executio n of sbi. [ example ] $20h=0 5h , i=04 h : ldi 20h,0 5h ;$20h=0 5h,a c=05 h,cy =0 s b i 2 0 h , 0 4 h ; a c = 0 1 h , $ 2 0 h = 0 5 h , c y = 1 : [ example ] $20h=0 2h , i=07 h : ldi 20h,0 2h ;$20h=0 2h,a c=02 h,cy =0 sbi 20h,0 7h ;ac=0b h,$2 0h=02 h ,cy = 0 : i n st ru ct ion: s b i m fun c tion: to subtra ct imme diate dat a i from the value of dat a memo ry , and then save the result in both accu mulato r and the dat a memo ry format: instru ction code: ca rry flag: ope r ation: sbim m x d i 0101 1 iiii xxx xxxx cy ac , mx ? mx - i explanation system mov e ment of sb im is to subt ra ct immedi a t e dat a i fro m the value of dat a memo ry , and save the re sult in b o th a c cumul a tor a nd the dat a memo ry . when exe c utin g sbim, i f the value of dat a memory is less than the immediate dat a , a ?borrow? will t a ke place and cy will be set to 0. on the cont rary , if the va lue of dat a memo ry is g r eater th an th e immediate d a t a, borrow wil l not happe n and cy will b e 1. theref ore we ca n d e cid e by the value of cy wheth e r the r e is a borro w a f ter execution of sbim. [ example ] $20h=0 5h , i=04 h : ldi 20h,0 5h ;$20h=0 5h,a c=05 h s b i m 2 0 h , 0 4 h ; a c = 0 1 h , $ 2 0 h = 0 1 h , c y = 1 : v e r1 .0 18/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n [ example ] $20h=0 2h , i=07 h : ldi 20h,0 2h ;$20h=0 2h,a c=02 h sbim 20h,0 7h ;ac=0b h,$2 0h=0b h ,cy=0 : i n st ru ct ion: d a s fun c tion: to adjust the valu e of dat a me mory to deci m al af ter subt ra ction, a nd save the result in accu mulator a nd the dat a m e mo ry format: instru ction code: ca rry flag: ope r ation: das mx 1 1 001 1010 xxx x xxx cy ac ; mx ? dec imal , adjust ac for sub explanation instru ction das act s by adjustin g the value of dat a memo ry to decimal af te r su btra ction and saving t he re sult to b o th accum u la tor and th e d a t a mem o ry . it s adju s ting method i s if the value of d a t a memory is greate r than 9 or if cy=0, then add 0ah to the dat a memory an d s e t cy to 0. [ example ] 05h - 06h , and do da s adjustme n t : ldi 20h,0 6h ;ac=0 6 h,$ 2 0 h =06 h ldi 21h,0 5h ;ac=0 5 h,$ 2 1 h =05 h l d a 2 0 h , 0 ; a c = 0 6 h s u b 2 1 h , 0 ; a c = 0 f h , c y = 0 d as 2 1 h ;ac = 09 h , $2 1h =0 9h ,c y= 0 : v e r1 .0 19/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n 1-5 instructions for logic operation logi c instructions are essential to sy stem structu r e. SH6610 se rie s mcu provid e som e commo n logi c in structions. no w i ? m goin g to e x plain to you one by one in most det ail s , and a ssi st m y explanation with simpl e exampl e s , so that you can quickly unde rst and a c tio n theory of ea ch instructio n. i n st ru ct ion: a nd fun c tion: to do logic a nd o peration with the values of dat a m e mo ry and accu mu lator , an d then save t he re sult in the ac cumulator format: instru ction code: ca rry flag: ope r ation: and mx d bbb 001 10 1bbb xxx x xxx not af fec t ed ac ? mx & ac explanation in and operation, the resul t w ill be 1(t r ue) only if both of the two operands are 1(true). it s logic t a ble is as follo ws: logi c ope rati on t able fo r and a b a and b 0 0 0 0 1 0 1 0 0 1 1 1 however , in real inst ru ction the logic op era nd has 4 bit s rather than 1 bit. the inst r u ct io n a nd i s to and the valu es of dat a me mory with accum u lato r , a nd the re sult i s saved in the accum u lato r . [ example ] 06h & 05h : l d i 2 0 h , 0 1 1 0 b ; a c = 0 6 h , $ 2 0 h = 0 6 h ldi 21h,0 101b ;ac=0 5 h,$ 2 1 h =05 h and 20h,0 ;ac=0 100b,$ 20h=0 6h : v e r1 .0 20/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: a n d m fun c tion: to do logi c a nd ope ratio n with the values of dat a memo ry and accu mulato r , and then save the re su lt in both the accu mulator a nd the dat a mem o ry format: instru ction code: ca rry flag: ope r ation: andm mx d bbb 001 10 1bbb xxx x xxx not af fec t ed ac , mx ? mx & ac explanation system movement of the instru ctio n andm i s al most the sa me as and, but saving th e ope ration result in dat a me mory as well as in ac cumulator . [ example ] 01 10b & 0101b : ldi 20h,0 1 1 0b ;ac=0 1 1 0 b,$20h=0 1 1 0b ldi 21h,0 101b ;ac=0 101b,$ 21h=0 101b andm 20h,0 ;ac=0 100b,$ 20h=0 100b : i n st ru ct ion: a ndi m fun c tion: to do logi c a n d ope ration wit h the valu e o f dat a m e mo ry and imme dia t e dat a i, an d then save the re sult in both accumulato r and the dat a memo ry format: instru ction code: ca rry flag: ope r ation: andim mx d i 01 1 10 iiii xxx xxxx not af fec t ed ac , mx ? mx & i explanation system movement of the instru ction a ndim i s to chang e ope ra nd 2 (a ccum ulator) of instructio n and to imme d i ate dat a i. this in st ru cti on is in imme diate mod e , so the add re ss of dat a memory can only be set to bank 0( $000 h ~ $07fh ). the operation re sult is save d in both accu mulator a nd the dat a m e m o ry . [ example ] $20h=01 10b , i=001 1 b : l d i 2 0 h , 0 1 1 0 b ; a c = 0 6 h , $ 2 0 h = 0 1 1 0 b andim 20h,0 01 1b ;ac=0 010b,$ 20h=0 010b : v e r1 .0 21/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n progr amming tip andim it self has a spe c ial function ?ma s k? . when we ne ed to set a cert ain bi t to 0, we can c l ear this bit to 0 with andim like this : [ example ] clear bit 2 of $20h to 0 a n d i m 2 0 h d 10 1 1 b af ter execution: $20 h=x0xxb v e r1 .0 22/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: o r fun c tion: to do logic o r ope ration wit h the values of dat a mem o ry and accu mu lator , an d then save t he re sult in the ac cumulator format: instru ction code: ca rry flag: ope r ation: o r m x d b bb 00101 0bbb xxx xxxx not af fec t ed ac ? mx | ac explanation in or operation, the resul t will be 1(true) if either one of the two ope ran d s is 1(tru e ). it s logic t a ble is as follo ws: logi c ope rati on t able fo r o r a b a o r b 0 0 0 0 1 1 1 0 1 1 1 1 however , in real inst ruction the l o gic op erand also h a s 4bit s rath er tha n 1 bit. the instruction or is to o r the values of dat a memo ry with accu mul a tor , and th e re sult is save d in the accum u lator . [ example ] 0001b | 0100b : ldi 20h,0 001b ;$20h=0 001 b,ac=00 01b ldi 21h,0 100b ;$21h=0 100 b,ac=01 00b or 20h,0 ;$20h=0 001 b,ac=01 01b : i n st ru ct ion: o r m fun c tion: to do logic o r ope ration wit h the values of dat a mem o ry and accumul a tor , and the n save the result in both the accumulato r and the dat a memo ry format: instru ction code: ca rry flag: ope r ation: orm mx d bbb 00101 1bbb xxx xxxx not af fec t ed ac , mx ? mx | ac v e r1 .0 23/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n explanation system movement of the in stru ction o r m i s almo st the same a s o r , but saving t h e ope ration result in dat a me mory as well as in ac cumulator . [ example ] 0001b | 0100b : ldi 20h,0 001b ;$20h=0 001 b,ac=00 01b ldi 21h,0 100b ;$21h=0 100 b,ac=01 00b orm 20h,0 ;$20h=0 101 b,ac=01 01b : progr amming tip in program designi ng, if a cert ain bit of a variable need s to be set to 1 and the other bit s m u st not be a f fected, then this ca n be do ne by t he inst ru ction or. beca use any bit that has done or ope ration with 0 can ke ep it s origi nal value , while tho s e with 1 will ha ve the valu e of 1. ins t ruc t ion: orim fun c tion: to do logi c o r ope ration with the value o f dat a mem o ry and imme dia t e dat a i, an d then save the re sult in both accumulato r and the dat a memo ry format: instru ction code: ca rry flag: ope r ation: o r i m m x d i 01 101 iiii xxx xxxx not af fec t ed ac , mx ? mx | i explanation system movement of the instru ction orim is to do logic o r ope ration wit h the value o f dat a m e mo ry and imm edi ate da t a i, a nd save the re sult in b o th accum u lato r and th e dat a memo ry . this inst ru ction i s also in imm ediate mo de. [ exa mple ] set bit 3 of the value of $20h to 1 o r i m 2 0 h d 10 00b af ter exec ution: $20h = 1xxxb v e r1 .0 24/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: e o r fun c tion: to do logic excl usive or ope ra tion with the value s of dat a memo ry and accumulato r , and the n sav e the re sult in the ac cumulator format: instru ction code: ca rry flag: ope r ation: eor mx d bbb 00100 0bbb xxx xxxx not af fec t ed ac ? mx ^ ac explanation system movement of eor is to do logic exclu s ive or operation with the value s of dat a memo ry and accum u lato r , and save the re sult in the a c cumul a tor . eor is usuall y referred to as exclusive or, becau se the operati on resu lt will be 1 only if the two opera n d s have dif f erent valu es; othe rwi s e the result will be 0. the logic t a ble for eor is a s foll ows: logi c ope rati on t able fo r eor a b a e o r b 0 0 0 0 1 1 1 0 1 1 1 0 [ example ] 001 1b ^ 0101b : ldi 20h,0 01 1b ;$20h=0 01 1b ,ac=0 0 1 1 b ldi 21h,0 101b ;$21h=0 101 b,ac=01 01b eor 20h,0 ;$20h=0 01 1b ,ac=0 1 1 0 b : v e r1 .0 25/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: e o rm fun c tion: to do lo gic excl usive or op eratio n with the valu es of dat a mem o ry and accu mu lator , and the n save the re sult in both the accum u lator a n d the dat a m e mory format: instru ction code: ca rry flag: ope r ation: e o r m m x d bbb 00100 1bbb xxx xxxx not af fec t ed ac , mx ? mx ^ ac explanation system movement of the instructio n eorm is almost the sa me as eor, i.e. doing eor action with th e values of dat a memo ry and a c cumul a tor , but saving the op erat ion re sult in th e dat a mem o ry as well as in t he accu mulat o r . [ example ] 001 1b ^ 0101 b : ldi 20h,0 01 1b ;$20h=0 01 1b ,ac=0 0 1 1 b ldi 21h,0 101b ;$21h=0 101 b,ac=01 01b eorm 20h,0 ;$20h=0 1 1 0b ,ac=0 1 1 0 b : i n st ru ct ion: e o ri m fun c tion: to do logi c eo r operation with the value o f dat a me mory and imme dia t e dat a i, an d then save the re sult in both the accumul a tor and the dat a memo ry format: instru ction code: ca rry flag: ope r ation: eorim mx d i 01 100 iiii xxx xxxx not af fec t ed ac , mx ? mx ^ i explanation operan d 2 of the instructi on eo rim shoul d be immediate dat a. this instruction i s t o d o logic eor op eratio n with the value of d a t a memory and immedi ate dat a i, and to save th e re sult in both the accum u la tor and the d a t a memo ry . [ example ] $20h=0 01 1b d i= 01 01 b : ldi 20h,0 011b ;$20h=0 011 b,ac=00 11b eorim 20h,0 101b ;$20h=0 1 1 0b ,ac=0 1 1 0 b : v e r1 .0 26/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n progr amming tip af ter readi ng all of the logic inst ru ctions, yo u may wond er why there ha ven ? t the ?n o t ? inst r u ct io n (in v e r se)? if there ha ve n ? t s u ch an in stru ction in sh66 10 se rie s , wh at can i do? don ? t worry , prog ram m i ng t i p is going to tell you how to use other instru ction s to perform th e not fu nctio n . the opera t ion of no t i s to chan ge 0 ? 1 or 1 ? 0 on e a ch bit. here, the eor in stru ct ion can h e lp us to g e t the in ve rsed valu e b y doi ng eo r op eratio n wit h imm ediate da t a (0f h ) and the va riabl e that wa nt s to be do ne not . y o u ? ll underst an d cle a rl y af ter readi ng the followin g exam ple. [ example ] do the ope ratio n not with the value of $2 0h (1 1 00b) e o r i m 2 0 h , 0 f h execution re su lt: $20h=00 1 1b,ac=0 0 1 1 b 1-6 instruc t ions for f l o w contr o l i n st ru ct ion: j m p fun c tion: to jump to a de si gnated a d d r e ss to exe c ute prog ram format: instru ction code: ca rry flag: ope r ation: j m p x 1 1 10p xxxx xxx xxxx not af fec t ed pc ? x(in clu de p) explanation instru ctio n jmp jump s to a desi gnated a ddress to execute prog ram. howeve r , addressi ng cap a bility of sh66 10 seri es cpu is limit ed to 4k wo rds (000 0h~0 fffh), so th e jumpin g ran g e of jmp ca n only rea c h 4 k (0fff h). addre s s bey ond 4k shall be re ached b y swit chin g ba nks. there? s det aile d explanatio n in later ch apters for how to swit ch ban ks. the jmp in st ru ct ion is simi lar t o t he go t o inst r u ct io n in basic program. [ exampl e ] pc= 4 0h , j u mp to 0e00h j m p 0 e 0 0 h e x e c u t i o n r e s u l t : p c = 0 e 0 0 h v e r1 .0 27/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n [ exampl e ] j u mp to lable jmp loop ;jump to loop (pc ? 034 0h ) : o r g 0 3 4 0 h l o o p : n o p n o p : i n st ru ct ion: b a z fun c tion: if ac equ als 0 , then go to a desi gnate d add re ss t o execute pro g r am; othe rwi s e co ntinue to execute the next line format: instru ction code: ca rry flag: ope r ation: baz x 10010 xxxx xxx xxxx not af fec t ed pc ? x , if a c =0 explanation if the val ue of ac is 0, then af ter exec uting the baz instruction, pc will go to th e desi gnate d a ddress x to execute p r o g r am, t he ra n ge of x bein g from $0 00 h to $7ff h . i t contin ue s to execute the n e xt line if the value of ac is 1 [ example ] if ($20 h=$2 0h-1 )=00 h then goto i n c21 h ldi 20h,0 fh ;$20h=0 fh dec20 h : sbim 20h,0 1h ;ac,$20 h ? $20 h -1 b a z i n c 2 1 h ; i f a c = 0 j u m p t o inc21 h j m p d e c 2 0 h ; e l s e j u m p t o dec 2 0 h : inc21 h adim 21h,0 1h ;$21h+1 : i n st ru ct ion: b a 0 fun c tion: if bit 0 of ac is 1, then go to a desig nat ed add re ss t o execute prog ram; othe rwise co ntinue to execute th e next line format: instru ction code: ca rry flag: ope r ation: ba0 x 10100 xxxx xxx xxxx not af fec t ed pc ? x , if a c (0)= 1 v e r1 .0 28/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n explanation if bit 0 of ac is 1, th en af ter executing ba0 instructio n, pc will go to the designat ed add re ss x to execute program, the ra ng e of x bei ng f r om $ 000 h to $7ff h or f r o m $080 0 h to $0fff h. it contin ue s to execute the n e xt line if bit 0 of ac is 0 . [ example ] if $20 h(bit 0)=1 then goto inc21 h : ldi 20h,0 fh ;$20h=0 fh,a c=0f h dec20 h : bim 20h,0 1h ;$20h,a c ? $ 20h -1 b a 0 i n c 2 1 h ; i f a c ( b i t 0 ) = 1 , j u m p t o i n c 2 1 h j m p d e c 2 0 h ; e l s e j u m p t o d e c 2 0 h : inc2 1h: adim 21h,0 1h ;$21h,a c ? $ 21h+1 i n st ru ct ion: b a 1 fun c tion: if bit 1 of ac is 1, then g o to de sign a t ed add re ss to execute pro g r am; othe rwi s e co ntinue to execute the next line format: instru ction code: ca rry flag: ope r ation: ba1 x 10101 xxxx xxx xxxx not af fec t ed pc ? x , if a c (1)= 1 explanation if bit 1 of ac is 1, th en af ter executing ba1 instructio n, pc will go to the designat ed add re ss x to execute program, the ra ng e of x bei ng f r om $ 000 h to $7ff h or f r o m $080 0 h to $0fff h. it contin ue s to execute the n e xt line if bit 1 of ac is 0 . [ example ] if $20h(bit 1)=1 then goto inc21 h : ldi 20h,0 fh ;$20h=0 fh,a c=0f h dec20 h : sbim 20h,0 1h ;$20h,a c ? $ 20h -1 b a 1 i n c 2 1 h ;ifac(bit1)= 1,jump to inc21h j m p d e c 2 0 h ; e l s e j u m p t o d e c 2 0 h : inc21 h : adim 21h,0 1h ;$21h,a c ? $ 21h+1 : v e r1 .0 29/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: b a 2 fun c tion: if bit 2 of ac is 1, then g o to desig na ted add re ss to execute prog ram; othe rwise co ntinue to execute th e next line format: instru ction code: ca rry flag: ope r ation: ba2 x 101 10 xxxx x xx x xxx not af fec t ed pc ? x , if a c (2)= 1 explanation if bit 2 of ac is 1, th en af ter executing ba2 instructio n, pc will go to the designat ed add re ss x to execute p r og ram, the rang e of x being from $0 00 h to $7ffh o r fro m $0800 h or $0fff h. it contin ue s to execute the n e xt line if bit 1 of ac is 0 . [ example ] if $20 h(bit 2)=1 then goto inc21 h : ldi 20h,0 fh ;$20h=0 fh,a c=0f h dec20 h : sbim 20h,0 1h ;$20h,a c ? $ 20h -1 ba2 inc21h ;if ac(bit2)= 1 , jump to inc21h j m p d e c 2 0 h ; e l s e j u m p t o d e c 2 0 h : inc2 1h: adim 21h,0 1h ;$21h,a c ? $ 21h+1 : i n st ru ct ion: b a 3 fun c tion: if bit 3 of ac is 1, then g o to de sign a t ed add re ss to execute pro g r am; othe rwi s e co ntinue to execute the next line format: instru ction code: ca rry flag: ope r ation: ba3 x 101 1 1 xxxx x xx x xxx not af fec t ed pc ? x , if a c (3)= 1 explanation if bit 3 of ac is 1, th en af ter executing ba3 instruct ion, pc will go to the designat ed add re ss x to execute program, the ra ng e of x bei ng f r om $ 000 h to $7ff h or f r o m $080 0 h to $0fff h. it contin ue s to execute the n e xt line if bit 1 of ac is 0 . v e r1 .0 30/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n [ example ] if $20 h(bit 3)=1 then goto inc21 h : ldi 20h,0 fh ;$20h=0 fh,a c=0f h dec20 h sbim 20h,0 1h ;$20h,a c ? $ 20h -1 b a 3 i n c 2 1 h ; i f a c ( b i t 3 ) = 1 , j u m p t o i n c 2 1 h j m p d e c 2 0 h ; e l s e j u m p t o d e c 2 0 h : inc2 1h: adim 21h,0 1h ;$21h,a c ? $ 21h+1 : i n st ru ct ion: b c fun c tion: if cy is 1, then go to desi gnated a ddre ss to execute pro g ra m; otherwise co ntin ue to execute the next line format: instru ction code: ca rry flag: ope r ation: b c x 1001 1 xxxx x xx x xxx cy pc ? x , if cy= 1 explanation i f c y i s 1 , then af ter ex ecutin g in stru ction bc, pc will go to the desi gnate d add re ss x t o execute p r og ram, the ran ge of x being from $000 h to $7ffh. it continue s to execute the next line if cy is n o t 1. the in structi on bc is of ten u s e d af te r ad dition or subt ra ction t o deci de wh eth e r there is a carry or bo rro w . y ou sho u l d espe cially not e that for addition, cy is set to 1 when there i s a ca rry , while for subt ra ction cy is set to1 when the r e i s n? t any borro w . therefore you sho u ld be careful wh e n d ealing with program flows. [ example ] if cy=1 then g o to inc2 0h : ldi 20h,0 fh ;$20h=0 fh,a c=0f h,cy=0 inc2 0h : sbim 20h,0 1h ;$20h,a c ? $ 20h -1 bc inc20h ;if cy= 1 ,jump to inc20h : v e r1 .0 31/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: t j mp fun c tion:uncondition ally go to the address com p o s ed by (pc1 1 ~ pc8), tbr and ac value to execute pro g ra m format: instru ction code: ca rry flag: ope r ation: tjmp 11 11 0 1 11 1 1 1 1 11 11 not af fec t ed explanation the de stinati on add re ss for instructio n tjmp is co mpo s ed by pc? s bit 8 ~ bit 1 1 , tb r and ac valu e (plea s e refer to instru ction r t nw for p r ogram examp l e). example: pc=30 0 h, tbr=01h, ac=0 2 h , then the ru le for com p o s ing a de stinat ion addres s is as follows : add re ss=pc(bit8~bit 1 1 ) tbr ac if: p c = 3 00h t b r = 0 1 h a c = 0 2 h then the d e st ination ad dre ss i s : 3 1 2 h instru ction: call fun c tion: to call a sub p rogram format: instru ction code: ca rry flag: ope r ation: call x 1 1 000 xxxx x xx x xxx not af fec t ed st ? cy ;pc , pc ? x (not includ e p) explanation instruction call is u s e d to call a su bprogram. f i r s t i t s a v e s the values of cy and pc+ 1 to st ack for returning to the calling program, then g o e s to the d esi g nated a dd r e s s x ( $ 000 0h ~ $07ff h or $0800 h ~ $0fff h) to execute p r og ra m. instructi ons r t nw o r r t ni ca n be use d to retu rn to the calli ng prog ram. when usi n g call to call a su bprog ra m, you sho u l d esp e ci ally note how many layers of st a ck have al rea dy been use d , becau se s h 66 10 se rie s only provide 4-laye r st a c ks. if more than 4 laye rs are used, serious erro r will be o c cu rre d when returning to the calling program! v e r1 .0 32/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: r t n w func tion: to return to the calling program, h ? tb r, l ? ac format: instru ction code: ca rry flag: ope r ation: rt n w h d l 1 10 10 00 0h h hh 1 1 1 1 not af fec t ed pc ? st , tbr ? hh hh , ac ? 111 1 explanation r t nw is an instructio n to get dat a from st ack to pc for returnin g to the calling pro g ra m , and at the same time put the value of h into tb r and the value of l into ac. this instructi o n is of ten u s ed to get st ation a ry dat a. [ example ] t o get dat a from rom a d d r e ss 3 0 2 h tbr equ oeh temp equ 20h : : 001a ldi tbr d 0 0 h ;put index value (hi gh nib b l e ) 0 into tbr. 001b ldi temp d 02 ;put index value (lo w nib b l e ) 2 into ac 001 c call 300 h ;call su bp rog r am. 001 d : : : o r g 3 0 0 h 0300 tjmp ; get destination add re ss $ 030 2h a c cording to (pc1 1~ pc8), t br,ac 0301 r t nw 00h,0 1h 0302 r t nw 00h,0 2h ;return to mai n pro g ra m, h ? tb r,l ? ac 0303 r t nw 04h,0 5h 0304 r t nw 09h,0 8h 0305 : i n st ru ct ion: r t ni func tion: to return from interrupt or sub p ro gra m format: instru ction code: ca rry flag: ope r ation: rt n i 1 10 10 10 00 0 00 000 0 cy cy ;pc ? st v e r1 .0 33/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n explanation instru ction r t ni is mainly used for retu rnin g from interrupt or sub p ro gra m . it fills cy and pc with values of st ack (cy an d returni ng add re ss) wh en returning. what? s the dif f eren ce b e twee n r t ni a nd r t nw? w e can find that whe n ret u rni ng by r t nw , only the returning ad d r e ss in the st ack is fetche d into pc, but cy value is not fetched. and r tnw fetche s anoth e r two valu es (h ? tbr, l ? ac ), whi c h r t ni doe s n o t do. therefore you ca n cho o se from the two in stru ction s acco rdi ng to your ne eds. [ example ] t o exchang e two numb e rs 000e 1 tbr equ 0eh 0020 2 regx equ 20h 0021 3 regy equ 21h 0022 4 temp equ 22h 5 ;********** ** ******** * 0005 780e 12 reset : ldi temp ,00h ;set temp=00h 0006 7 920 13 ldi regx,02 h ;set regx =0 2 h 0007 7a 21 14 ldi regy ,04h ;set regy =0 4 h 0008 c00a 15 call sw apxy ; ;call subprogram 17 ;******** *** ******** *** 000a 3820 18 sw apxy lda regx,00h ;ac=02h 000b 3 c 22 19 st a temp ,00h ;temp=02 h 000 c 382 1 20 lda regy ,00h ;ac=0 4h 000 d 3c20 21 st a regx,00 h ;regx=0 4 h 000e 38 22 22 lda temp ,00h ;ac=0 2 h 000f 3 c 2 1 23 st a regy ,00h ;reg2 = 02 h 0010 d40 0 24 r t ni ;return to mai n pro g ra m v e r1 .0 34/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n i n st ru ct ion: h a l t fun c tion: cp u to be halt from wo rki n g format: instru ction code: ca rry flag: ope r ation: hal t 1 10 1 1 0 000 0 00 000 0 not af fec t ed no explanation af ter executing inst ruction ha l t , cpu will be ha lt while it s surrounding circuit (counter , oscillation circuit) continues working. the instruct ion ha l t is usu ally use d to stop cp u temporarily in orde r to save powe r . in hal t mode, when any of t he system inte rru pt s o c curs, cpu will be relea s ed from hal t m ode a nd co ntinue t o wo rk. [ example ] halt pro g ram, to ena ble port b interrupt to wak e up the program iex equ 00h ;interru pt ena ble re giste r irq equ 01h ;interru pt req u ire flag por t b equ 09h ;i/o port b : ldi por t b,0f h ;set port b = ? high ? ldi iex,0001b ;enable p o rt i n terrupt ldi irq,00 h ;clea r interrup t require flag h a l t ; s y s t e m c p u h a l t n o p : i n st ru ct ion: s t op fun c tion: to stop the whol e chip (i ncl udin g oscillation cir c uit) format: instru ction code: ca rry flag: ope r ation: st op 1 10 1 1 1 000 0 00 000 0 not af fec t ed no explanation executing inst ru ct ion st op will stop the whole chip from working, incl uding oscillation c i rc uit. only por t interrupt and external interrupt can relea s e cpu from s t op mode, so you must en able an inte rrupt befo r e e n terin g st op mode , otherwise the system can?t be wak ed up from st op mode. v e r1 .0 35/ 36
s h 6 6 1 0 i n s t r u c t i o n s i n t r o d u c t i o n [ example ] stop prog ra m, to enable port b interrupt to wa ke up the progra m iex equ 00h ;interru pt ena ble re giste r irq equ 01h ;interru pt req u ire flag p o r t b e q u 0 9 h ; i / o p o r t b : ldi por t b,0f h ;set port b = ? high ? ldi iex,0001b ;enable p o rt i n terrupt ldi irq,00 h ;clea r interrup t require flag s t o p ; a l l s y s t e m i s ? s t o p ? n o p : i n st ru ct ion: n o p fun c tion: to do nothing format: instru ction code: ca rry flag: ope r ation: nop 11 11 11 11 11 1 11 11 not af fec t ed no explanation instruction nop mea n s doing n o thin g in it s in stru ction cycle a nd it is of ten use d for tim e delay . beca use it does n o thing wh en executin g, you don?t worry if it will af fect any current st at u s . v e r1 .0 36/ 36

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