在程序集x86中使用CALL,RET

Question

考虑其使用以下简单汇编代码push,pop和call

CDSEG SEGMENT
MAIN PROC FAR
    ASSUME CS:CDSEG,DS:DTSEG
    MOV AX,DTSEG
    MOV DS,AX

    MOV AX,1010H        ; (1)
    PUSH AX             ; (2)
    CALL FOO            ; (3)
    ADD AX,2            ; (7)
MAIN ENDP   
FOO PROC   
    POP AX              ; (4)
    ADD AX,1            ; (5)
    RET                 ; (6)
FOO ENDP
END MAIN

我期待的是,看到

(1) ax = 1010h, stack=????
(2) ax = 1010h, stack=1010h
(3) 
(4) ax = 1010h, stack=????
(5) ax = 1011h
(6)
(7) ax = 1012h

但是,模拟器显示

(1) ax = 1010h, stack=????
(2) ax = 1010h, stack=1010h
(3) 
(4) ax = 000Ch, stack=????
(5) ax = 000Dh
(6) The control never goes back to the main proc
(7) ???

什么是错？

Answer 1

CALL推送堆栈上的返回地址,这是允许RET返回该地址的地址.当你POP AX在FOO弹出堆栈的返回地址时.当你输入时,你PUSH之前编辑的值CALL不在堆栈的顶部FOO,而是在[SP+2].