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Old 08 January 2018, 23:44   #1
dalton
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vbcc 0.9e signed multiplication issue?

I'm struggling with the most basic of things now it seems.

This function (signed multiplication)

Code:
int foo(int a, int b)
{
	return a*b;
}
compiles to this (unsigned multiplication)

Code:
; stacksize=0
	near	code
	opt	0
	opt	NQLPSMRBT
	public	_foo
	cnop	0,4
_foo
	movem.l	l132,-(a7)
	move.l	(4+l134,a7),d0
	move.l	(8+l134,a7),d1
	move.l	d0,d2
	move.l	d1,d3
	swap	d2
	swap	d3
	mulu.w	d1,d2
	mulu.w	d0,d3
	mulu.w	d1,d0
	add.w	d3,d2
	swap	d2
	clr.w	d2
	add.l	d2,d0
l130
l132	reg	d2/d3
	movem.l	(a7)+,d2/d3
l134	equ	8
	rts
My command line looks like this:
Code:
devkit/tools/win32/vbccm68k.exe -I. -Idevkit/vbcc-targets/m68k-amigaos/include -Idevkit/include -c99 -quiet -D__VBCC__ -D__AMIGA__ -D__MINSTART__ -DOVERSCAN=1 -cpu=68000  -O=0 -maxoptpasses=65535 -dontwarn=72     -sc  -DNDEBUG  -o=main._s main.c
I must be doing something wrong?
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Old 09 January 2018, 00:09   #2
Samurai_Crow
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A 32 bit multiply on a 16 bit CPU?
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Old 09 January 2018, 01:46   #3
robinsonb5
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Quote:
Originally Posted by dalton View Post
I must be doing something wrong?
Are you getting incorrect results?
You're asking for a pair of 32-bit values to be multiplied, and for the lower 32-bits of the result (which would be 64-bit) to be returned, and that's what the code is doing. It doesn't matter that the code is using mulu for this - I think the difference in results between signed and unsigned multiplication will be in the upper 32-bits of the result, which you're discarding.
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Old 09 January 2018, 09:47   #4
dalton
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Quote:
Originally Posted by robinsonb5 View Post
Are you getting incorrect results?
You're asking for a pair of 32-bit values to be multiplied, and for the lower 32-bits of the result (which would be 64-bit) to be returned, and that's what the code is doing. It doesn't matter that the code is using mulu for this - I think the difference in results between signed and unsigned multiplication will be in the upper 32-bits of the result, which you're discarding.
You're right, thanks! It's true that signed and unsigned multiplication are equivalent in this case. I believed I had it figured out last night but it turns out I did not put enough thought into it.

I'm having weird sign-related errors in my program, but it must be related to something other than multiplication.
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