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Programming with NASM / Re: Why do MUL and IMUL give the same output?
« Last post by fredericopissarra on October 13, 2019, 10:58:53 AM »I still get the same output value of 0x8000 stored in AX. However, this is invalid, since both inputs are positive, and therefore the output is clearly also positive, but positive 0x8000 is an INVALID value for a 16bit signed integer. Why does the CPU actually complete this calculation, instead of throwing an error (which then would need to be handled by SEH, structured exception handling, in the program)?You are forgetting that MUL/IMUL results, in this case, are 32 bits values stored in DX:AX.
It seems that with the CPU ignoring this error condition, MUL and IMUL operators are actually identical.
Using IMUL, 0x4000 times 2 is 0x00008000 (32 bits).
MUL/IMUL doesn't raise exceptions, they affect the flags (CF and OF - they have the same meaning: flagging overflow condition). Here's a simple test:
Code: [Select]
; mul.asm
bits 64
section .text
; int testmul( int *carry );
global testmul
testmul:
mov ecx,0x40000000
mov eax,2
imul ecx
mov ecx,0
setc cl
mov [rdi],ecx
retCode: [Select]
/* test.c */
#include <stdio.h>
extern int testmul( int * );
int main( void )
{
int r, c;
r = testmul(&c);
printf( "result=%#x, carry=%d\n", r, c );
}This will result in:Code: [Select]
result=0x80000000, carry=1As it should!