Re: [9fans] fp: stack underflow error

[Charles Forsyth <forsyth@terzarima.net>]

[-- Attachment #1: Type: text/plain, Size: 750 bytes --]

that produced invalid operation when i tried it,
not the stack underflow of the previous example.
the result won't fit in a signed integer, and unsigned
integer isn't supported by the hardware, so you get a trap
as documented.  in fact,
the compiler could and probably should compensate by
converting to and from 64-bit signed integers,
which would also eliminate the test-and-skip mentioned earlier.
i think 64-bit is supported, since it's listed with an opcode,
although the textual description then claims `word- or short-integer format'
only.  it's an Intel handbook.  ``beware DEC manuals''

the stack underflow is more likely to be a compiler problem.
(stack overflow on the other hand is quite likely to be a missing prototype.)

[-- Attachment #2: Type: message/rfc822, Size: 2879 bytes --]

From: rog@vitanuova.com
To: 9fans@cse.psu.edu
Subject: Re: [9fans] fp: stack underflow error
Date: Thu, 1 Jul 2004 20:24:23 +0100
Message-ID: <d66164ffbf07b27c99685ea41599f1a3@vitanuova.com>

> I encountered a "sys: fp: stack underflow" error in my program.

looks like a compiler bug.

here's a simpler version of the code that does the same thing.
who'd've thought that a2b() would generate such
convoluted code? what's that comparison for?

	TEXT	a2b+0(SB),0,$12
	MOVL	tb+4(FP),CX
	SUBL	fb+0(FP),CX
	MOVL	CX,.safe+-4(SP)
	FMOVL	.safe+-4(SP),F0
	CMPL	CX,$0
	JGE	,2(PC)
	FADDD	$(4.29496729600000000e+09),F0
	FMULD	dis+8(FP),F0
	FSTCW	,.safe+-8(SP)
	MOVW	$3967,.safe+-6(SP)
	FLDCW	.safe+-6(SP),
	FMOVLP	F0,.safe+-4(SP)
	FLDCW	.safe+-8(SP),
	MOVL	.safe+-4(SP),AX
	RET	,

#include <u.h>
#include <libc.h>
int
a2b(int fb, uint tb, double dis)
{
	return (tb - fb)*dis;
}

void
main(void)
{
	a2b(1, 0, 65);
	a2b(1, 0, 66);
}