匯編語(yǔ)言編程題

1、1、在屏幕上輸出：解：codesegmentassumecs:codestart:movbx9rotate:movdlmovcxbxprint:movah02hint21hloopprintmovdl0ahmovah02hint21hmovdl0chmovah02hint21hsubbx2jnsrotatemovah4chint21hcodeendsendstart2、從鍵盤上接收一個(gè)字符，如果為小寫字母則顯示其對(duì)應(yīng)的大寫字母。解：co

2、desegmentassumecs:codestart:movah01hint21hcmpalajlexitcmpalzjgexitmovdlalsubdl20hmov21hexit:movah4chint21hcodeendsendstart(題1的)思路：令cx=9dl=，循環(huán)輸出loopdl并且令cx自減2；使用換行語(yǔ)句；直到cx小于0，退出。此處注意注意dl的值是否改變，以免輸出的不是’’。3、鍵盤接收一個(gè)小寫字母，然后找出它的

3、前導(dǎo)字母和后續(xù)字母，按順序顯示這三個(gè)字符。解：codesegmentassumecs:codestart:movah01hint21hmovdlaldecdlcmpalajneprint1movdlzprint1:movah02hint21hmovdlalmovah02hint21hmovdlalincdlcmpalzjneprint2movdlaprint2:movah02hint21hmovah4chint21hcodeendsen

4、dstart7、設(shè)首地址為BUF的存儲(chǔ)單元，存放10個(gè)字?jǐn)?shù)據(jù)，求出它們的和，并將結(jié)果存入變量sum中。解：datasegmentbufdw10dup(2)sumdwdataendscodesegmentassumecs:codeds:datastart:movaxdatamovdsaxmovcx10movax0leabxbufrotate:addax[bx]addbx2looprotatemovsumaxmovah4chint21hco

5、deendsendstart8、在BUF1中和BUF2兩個(gè)數(shù)據(jù)區(qū)中各定義10個(gè)帶符號(hào)字符數(shù)據(jù)，將它們的對(duì)應(yīng)項(xiàng)的絕對(duì)值取和，并將結(jié)果存入以sum為首地址的數(shù)據(jù)區(qū)中。(如BUF1的第一項(xiàng)和BUF2的第一項(xiàng)的絕對(duì)值之和存入sum的地址中)解：datasegmentbuf1dw562413310516441421buf1dw45241966125274217sumdw10dup()dataendcodesegmentassumecs:coded