3.16 解：mov ah,1；只允許輸入小寫字母int 21hsub al,20h；轉換為大寫字母mov dl,almov ah,2int 21h；顯示3.18 解：mov ax, bufXcmp ax, bufYjae donemov ax, bufYdone:mov bufZ, ax3.19 解：.model small.stack.databufXdw -7signXdb ?.code.startupcmp bufX,0;test bufX,80hjl next;jnz nextmov signX,0jmp donenext:mov signX,-1done:.exit 0end3.20 解：mov dl,2mov ax,bufXcmp ax,bufYje next1dec dlnext1:cmp ax,bufZje next2dec dlnext2:mov ah,2int 21h3.22 編制程序完成12H、45H、0F3H、6AH、20H、0FEH、90H、0C8H、57H和34H等10個字節數據之和，并將結果存入字節變量SUM中（不考慮溢出和進位）。；wjxt322.asm.model small.stack.datab_datadb 12h,45h,0f3h,6ah,20h,0feh,90h,0c8h,57h,34h；原始數據numequ 10；數據個數sumdb ?；預留結果單元.code.startupxor si, si；位移量清零xor al, al；取第一個數mov cx, num；累加次數again:add al, b_datasi；累加inc si；指向下一個數loop again；如未完，繼續累加mov sum, al；完了，存結果.exit 0end3.30 解：lucaseprocpush bxmov bx,offset stringcmp al,0je case0cmp al,1jz case1cmp al,2jz case2jmp donecase0:cmp byte ptr bx,0je donecmp byte ptr bx,Ajb next0cmp byte ptr bx,Zja next0add byte ptr bx,20hnext0:inc bxjmp case0case1:cmp byte ptr bx,0je donecmp byte ptr bx,ajb next1cmp byte ptr bx,zja next1sub byte ptr bx,20hnext1:inc bxjmp case1case2:cmp byte ptr bx,0je donecmp byte ptr bx,Ajb next2cmp byte ptr bx,Zja next20add byte ptr bx,20hjmp next2next20:cmp byte ptr bx,ajb next2cmp byte ptr bx,zja next