文档介绍：第二次实验1、编写一个程序,把AL寄存器中的两位十六进制数显示出来。程序如下CODESEGMENT;段开始ASSUMECS:CODE;把代码段和寄存器CSSTART:MOVAL,3EHMOVBL,AL;把AL的数给BLMOVDL,AL;把AL的数给DLMOVCL,4;把4存入CLSHRDL,CL;DL右移4位,变成03HCMPDL,9;DL内的数和9比较JBENEXT1;小于等于9则进入nextlADDDL,7;大于9则DL=DL+7NEXT1:ADDDL,30H;把DL中的数转换成ASCII代码MOVAH,2;调用DOS中的2号中断命令,把DL中的数显示出来INT21H;显示高位ASCII码MOVDL,BLANDDL,OFH;和00001111逻辑与,变成OEHCMPDL,9JBENEXT2ADDDL,7NEXT2:ADDDL,30HMOVAH,2INT21H;显示低位ASCII码MOVAH,4CHINT21HCODEENDS;返回DOSENDSTART流程图r1J(1 )在Windows条件下直接运行运行把ANDDL,OFH改成SHLDL,CL,运行时错误,:3000cycles,Frameskip0tProgram:DEB…AX二0Z3EBX二0O3ECX=0OO4DX二0033SP二0000BP二0000SI二0000DI=00O0DS=0760ES=0760SS=076Fcs=e?70IP二0018NUUPElPLNZNAPENC0770:0018CDZ1IMT Z1-tAX二0Z3EBX二0O3ECX二0004DX=0033SP=FFFABP=0000SI二0000DI二0000DS=076OES=0760SS=076FCS=FOO0IP=14A0NUUPDIPLNZNAPENCF0OO:14A0FBSTI-tAX=GZ3EBX二O03ECX=O004DX=0G33SP=FFFABP二0000SI二0000DI=O00ODS二0760ES=076OSS=076FCS=F0O0IP=14A1NUUPElPLNZNAPENCFO00:14A1FE38??? [BX*SIJDS:0O3E二00-tAX二0Z33BX=0O3ECX=00O4DX=0G33SP=FFFABP二0000SI二0000DI二0000DS二0760ES=0760SS=076FCS=FO00IP=14A5NUUPElPLNZNAPENCF00O:14A5CFIRET-tAX二0233BX二0O3ECX=0OO4DX二0033SP二0000BP二0000SI二0000DI二0000DS=0760ES=0760SS=076Fcs=e?70IP=0O1ANUUPElPLNZNAPENC0770:0O1A8AD3MOU DL,BL其中IRET是中断返回指令,STI为置IF为零(2):3000cycles,Frameskip0fProgram:DOS…-:\>masmlicrosoft(R)(C)MicrosoftCorp1981-1985,'ross-referenceJnableto[.ASM!:disphex[]:[:[:openinputfile::\>masmlicrosoftopyright(R)(C)MicrosoftCorp1981-1985,-reference[.ASM]:[]:[]:[:51750+:3000cycles,Frameskip0fProgram:DOS…-::\>masmlicrosoft(R)(C)MicrosoftCorp1981-1985,.:ourcefilename