我正在使用tasm，我正试图在另一个程序中调用一个程序。但是，当程序从第二个程序返回时，不是返回到它被调用的点，而是将IP设置为0000.有人可以解释一下为什么会发生这种情况吗？ 以下是代码的一部分：
主要程序：
MAIN PROC FAR ; main proc MUST be FAR
ASSUME CS:COD1,DS:DATA1,SS:STCK
PUSH DS ;save DS on stack for OS return
XOR AX,AX ;
PUSH AX ;put 0 on stack for OS return
MOV AX,DATA1
MOV DS,AX ;load data Segement register
CALL PLAY
RET ; return to OS
MAIN ENDP
首先称为程序：
PLAY PROC NEAR
RPT:
MOV AH,00H ;read ch from stdin
INT 16H
CMP AH,48H
JE UP
CMP AH,4BH
JE LEFT
CMP AH,50H
JE DOWN
CMP AH,4DH
JE RIGHT
CMP AH,01H
JE ESCAPE
UP:
CALL MOVEUP
JMP RPT
DOWN:
CALL MOVEDOWN
JMP RPT
LEFT:
CALL MOVELEFT
JMP RPT
RIGHT:
CALL MOVERIGHT
JMP RPT
ESCAPE:
RET
PLAY ENDP
第二个称为程序的模型：
MOVEUP PROC NEAR ;tries to update position of player if down key pressed
PUSH DX
PUSH AX
PUSH SI ;save all general registers to stack
MOV AL,X ;compute index for current position
MOV DL,15
MUL DL
MOV DL,Y
MOV DH,0
ADD AX,DX
MOV SI,AX
PUSH SI ;save it for later
MOV DL, X ;current x coord
MOV DH, Y ;current y coord
SUB DL, 1
MOV NEWX,DL ;new x coord
MOV NEWY,DH ;new y coord
MOV AL,NEWX ;store new x coord in AL
MOV AH,0 ;make ah 0 to "extend" new x on 2B
CMP AL,0 ;check if new x in bounds (>=0)
JL NOT_OK_UP
MOV DL,15 ;compute index of position: index=15*x+y
MUL DL
MOV DL,NEWY
MOV DH,0
ADD AX,DX
MOV SI,AX
MOV DL,MAZE[SI] ;fetch element on position newx,newy
CMP DL,1 ;if wall, go to end of procedure(nop)
JE NOT_OK_UP
MOV DL,NEWX
MOV DH,NEWY
MOV X,DL ;update x
MOV Y,DH ;update y
MOV MAZE[SI],2 ;update position in maze
POP SI
MOV MAZE[SI],0 ;clean up old position
NOT_OK_UP:
POP SI
POP AX
POP DX
RET
MOVEUP ENDP