;ret and call assume cs:code stack segment db 8 dup(0) stack ends code segment mov ax,4c00h int 21h start: mov ax,stack mov ss,ax mov sp,8;stack top point mov bx,00h push bx ; mov ax,0 ; push ax ; mov bx,0 ret ;pop ip ;retf ;pop ip,pop cs code ends end start
事实上ret就是pop ip把stack里的东西取出来到ip,重新定向代码地址
stack不懂可以找我的另一个文章
标签:ip,mov,ret,指令,pop,ax,stack From: https://www.cnblogs.com/Frank-dev-blog/p/16927851.html