微機原理及應(yīng)用教材課后習(xí)題及答案-資料下載頁
2025-06-28 18:25本頁面
【正文】 30hcmp dl, 39hjbe htoendadd dl, 7htoend:rethtoasc endpcode endsend start3．28請按如下說明編寫子程序：；子程序功能：把用ASCII碼表示的兩位十進制數(shù)轉(zhuǎn)換位對應(yīng)二進制數(shù)；入口參數(shù)：DH=十位數(shù)的ASCII碼，DL=個位數(shù)的ASCII碼；出口參數(shù)：AL=對應(yīng)的二進制數(shù)答：stack segment stack db 100h dup(?)stack endscode segment 39。code39。assume cs:code, ss:stackstart:mov dx, 3932hcall asctohmov ax, 4c00hint 21hasctoh procpush cxsub dx, 3030hmov cl, 10mov al, dhmul cladd al, dlpop cxretasctoh endpcode endsend start3．29調(diào)用HTOASC子程序，編寫顯示一個字節(jié)的16進制數(shù)、后跟“H”的子程序。答：stack segment stack db 100h dup(?)stack endscode segment 39。code39。assume cs:code, ss:stackstart:mov ah, 07hint 21hcmp al, 1bhje donecall htoahjmp startdone:mov ax, 4c00hint 21hhtoah procpush axpush cxpush dxpush axmov dl, almov cl, 4shr dl, clcall htoascmov ah, 02hint 21hpop axmov dl, aland dl, 0fhcall htoascmov ah, 02hint 21hmov dl, 39。H39。mov ah, 02hint 21hpop dxpop cxpop axrethtoah endphtoasc procadd dl, 30hcmp dl, 39hjbe htoendadd dl, 7htoend:rethtoasc endpcode endsend start3．30寫一個子程序，根據(jù)入口參數(shù)AL=0、2，依次實現(xiàn)對大些字母轉(zhuǎn)換成小寫、小寫轉(zhuǎn)換成大寫或大小寫字母互換。欲轉(zhuǎn)換的字符串在string中，用0表示結(jié)束。答：data segmentstring db 39。abcDEfGh39。, 39。$39。data endsstack segment stack db 100 dup(?)stack endscode segment 39。code39。assume cs:code, ds:data, ss:stackstart:mov ax, datamov ds, axmov al, 2call changemov dx, offset stringmov ah, 09hint 21hmov al, 0call changemov dx, offset stringmov ah, 09hint 21hmov al, 1call changemov dx, offset stringmov ah, 09hint 21hmov ax, 4c00hint 21hchange proclea bx, stringcmp al, 0je cg1cmp al, 1je cg2cg3:cmp byte ptr [bx], 39。$39。je donecmp byte ptr [bx], 5ahja xtodadd byte ptr [bx], 20hjmp cg33xtod:sub byte ptr [bx], 20hcg33:inc bxjmp cg3cg2:cmp byte ptr [bx], 39。$39。je donecmp byte ptr [bx], 61hjb cg22sub byte ptr [bx], 20hcg22:inc bxjmp cg2cg1:cmp byte ptr [bx], 39。$39。je donecmp byte ptr [bx], 5ahja cg11add byte ptr [bx], 20hcg11:inc bxjmp cg1done:retchange endpcode endsend start3．34編寫一個計算字節(jié)校驗和的子程序。所謂“校驗和”是指不記進位的累加，常用于檢查信息的正確性。主程序提供入口參數(shù)，有數(shù)據(jù)個數(shù)和數(shù)據(jù)緩沖區(qū)的首地址。子程序回送求和結(jié)果這個出口參數(shù)。傳遞參數(shù)方法自定。答：data segmentcount = 10array db count dup(10h)check db ?data endsstack segment stack db 100 dup(?)stack endscode segment 39。code39。assume cs:code, ds:data, ss:stackstart:mov ax, datamov ds, axmov ax, countpush axmov ax, offset arraypush axcall sumcheckmov check, almov ax, 4c00hint 21hsumcheck procpush bpmov bp, sppush bxpush cxmov bx, [bp+4]mov cx, [bp+6]mov al, 0sum1:add al, [bx]inc bxloop sum1pop cxpop bxpop bpret 4sumcheck endpcode endsend start
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