How to fix "Unhandled exception" error in assembly? - function

Question

I've written a function that determines if a value is prime or not prime. But when I return from the function, it comes up with an error.
The error message is
Unhandled exception at 0x00000001 in Project.exe: 0xC0000005: Access violation executing location 0x00000001.
This function should return eax.
push ebp
mov ebp, esp
mov eax, [ebp+8] ; store input value
mov ecx, [ebp+8] ; store input value in counter
sub esp, 4 ; local variable
sub ecx, 1 ; avoid compare with itself
cmp eax, 3 ; compare with 1, 2, 3
jbe Prime
L1:
cmp ecx, 3 ; when count=3 to stop
je NotP
mov edx, 0 ; clear edx to save remainder
mov [esp-4], eax ; save input value
div ecx ; divide number
cmp edx, 0 ; check remainder
je NotP ; if remainder=0 then not prime
jmp Prime
loop L1
NotP:
mov eax, 0
push eax ; if delete this ilne still come up error
pop ebp
ret
Prime:
mov eax, 1
push eax ; if delete this ilne still come up error
pop ebp
ret
isPrime endp

Answer 1

mov [esp-4], eax ; save input value
If you plan on using the local variable that you reserved room for, then you have to write:
mov [esp], eax ; save input value
or alternatively write:
mov [ebp-4], eax ; save input value
A correct prolog/epilog would be:
push ebp
mov ebp, esp
sub esp, 4 ; local variable
mov eax, [ebp+8] ; store input value
...
NotP:
mov eax, 0
pop ebp ; remove local variable
pop ebp
ret
Prime:
mov eax, 1
pop ebp ; remove local variable
pop ebp
ret
isPrime endp
cmp edx, 0 ; check remainder
je NotP ; if remainder=0 then not prime
jmp Prime
loop L1
Finding the remainder not zero is not enough to conclude that the number is prime! More tests are needed. For now, that loop L1 instruction is never executed.
e.g. To test 15, your first division does 15 / 14 which yields a non-zero remainder but 15 isn't a prime number.
L1:
cmp ecx, 3 ; when count=3 to stop
je NotP
The top of the loop can't be correct either! Consider testing the number 7.
First division is 7 / 6 and has a remainder so the loop has to continue
Second division is 7 / 5 and has a remainder so the loop has to continue
Third division is 7 / 4 and has a remainder so the loop has to continue
You don't try any more divisions and conclude "not prime", yet 7 is definitively a prime number.