所以，我正在访问某个错误的地方，但我无法弄明白。我知道我的代码非常糟糕，所以请随意修复或指出错误。但是，我真正想知道的是如何修复此（标题）特定错误。 我只能理解MASM。 谢谢。
以下是我在MASM中的代码。
INCLUDE Irvine32.inc
.data
v1 dword 2
v2 dword 2
s1 byte "Please enter a number:",0
s3 byte "| The value will be recognized as non signed |",0Ah, 0
s4 byte "|By definition of Prime Number, number can't be negative|",0Ah, 0
.code
main proc
mov edx, offset s4
call writestring
mov edx, offset s3
call writestring
mov edx, offset s1
call writestring
call readdec
mov ecx, eax
dec ecx
dec ecx
L1:
push ecx
push v2
mov eax, v1
push eax
xor edx, edx
mov ebx, 2
div ebx
mov ecx, eax
call isPrime
pop v2
pop ecx
inc v1
cmp edx, 0
JE L4
LOOP L1
L4:
call writedec
dec ecx
JMP L1
exit
main endp
isPrime proc
L2:
pop eax
push eax
xor edx, edx
div v2
cmp edx, 0
JE L3
inc v2
LOOP L2
pop eax
mov edx, 0
ret
L3:
pop eax
mov edx, 1
ret <----------- ERROR OCCURS HERE.
isPrime endp
end main
好的，我解决了问题，现在它正在运行 代码在下面
INCLUDE Irvine32.inc
.data
v1 dword 2
v2 dword 2
s1 byte "Please enter a number:",0
s3 byte "| The value will be recognized as non signed |",0Ah, 0
s4 byte "|By definition of Prime Number, number can't be negative|",0Ah, 0
s5 byte "No Prime found.", 0Ah, 0
s6 byte "Primes found until the given number:", 0Ah, 0
.code
main proc
mov edx, offset s4
call writestring
mov edx, offset s3
call writestring
mov edx, offset s1
call writestring
call readdec
;if(eax < 2)
cmp eax, 2
JB L6
mov ecx, eax
dec ecx
dec ecx
;if(ecx < 1)
cmp ecx, 1
JB L7
L1:
push ecx
mov eax, v1
push eax
xor edx, edx
div v2
mov ecx, eax
pop eax
mov ebx, v2
call isPrime
pop ecx
inc v1
;if (edx == 0)
cmp edx, 0
JE L4
;if (eax == 2)
cmp eax, 2
JE L5
LOOP L1
exit
L4:
call writedec
mov al, ' '
call writechar
dec ecx
;if(ecx > 0)
cmp ecx, 0
JA L1
exit
L5:
mov edx, offset s6
call writestring
call writedec
mov al, ' '
call writechar
dec ecx
;if(ecx > 0)
cmp ecx, 0
JA L1
exit
L6:
mov edx, offset s5
call writestring
exit
L7:
mov edx, offset s6
call writestring
call writedec
mov al, ' '
call writechar
exit
main endp
COMMENT!
isPrime recieves eax and ebx. And, it returns edx as boolean, eax as number.
!
isPrime proc
L2:
push eax
xor edx, edx
div ebx
;if(edx == 0)
cmp edx, 0
JE L3
inc ebx
pop eax
LOOP L2
xor edx, edx
ret
L3:
pop eax
mov edx, 1
ret
isPrime endp
end main