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From:	Carl Fredrik Forsberg <Carl DOT Fredrik DOT Forsberg AT ngi DOT no>
To:	Eric Blake <eblake AT redhat DOT com>, "cygwin AT cygwin DOT com" <cygwin AT cygwin DOT com>
Subject:	RE: bug in lrint [was: FW: Printing long int in C program under cygwin64]
Date:	Wed, 24 May 2017 20:02:10 +0000
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Further to my first question, I also tried llrint() as below. It gives the same output as lrint(). This may not be a big surprise as long long int and long int have the same length.

#include <stdio.h>      /* printf */
#include <math.h>       /* lrint, llrint, rint */

int main ()
{
  char text[64];
  printf ( "type cast -1 = %li\n", (long int)-1 );
  printf ( "type cast lrint(-1.0) = %li\n", (long int)lrint(-1.0) );
  printf ( "rint(-1.0) = %f\n", rint(-1.0) );
  printf ( "lrint(-1.0) = %li\n", lrint(-1.0) );
  printf ( "llrint(-1.0) = %lli\n", llrint(-1.0) );
  printf ( "lrint(1.0) = %li\n", lrint(1.0) );
  printf ( "llrint(1.0) = %lli\n", llrint(1.0) );
  sprintf( text,"int -1 = %i", -1 );
  printf ( "Via sprintf: %s\n", text);
  printf ( "size of long long int: %zi\n", sizeof(long long int));
  printf ( "size of long int: %zi\n", sizeof(long int));
  printf ( "size of int: %zi\n", sizeof(int));
  return 0;

compiled by
gcc -Wall lrint_test.c -o lrint_test.exe

output:
type cast -1 = -1
type cast lrint(-1.0) = 4294967295
rint(-1.0) = -1.000000
lrint(-1.0) = 4294967295
llrint(-1.0) = 4294967295
lrint(1.0) = 1
llrint(1.0) = 1
Via sprintf: int -1 = -1
size of long long int: 8
size of long int: 8
size of int: 4
-----Original Message-----
From: cygwin-owner AT cygwin DOT com [mailto:cygwin-owner AT cygwin DOT com] On Behalf Of Eric Blake
Sent: 24 May 2017 18:57
To: cygwin AT cygwin DOT com
Subject: Re: bug in lrint [was: FW: Printing long int in C program under cygwin64]

On 05/24/2017 11:53 AM, Erik Bray wrote:

>>> dropping down to assembly; it could very well be that the assembly
>>> code is incorrectly truncating things at 32 bits (where it is just
>>> fine for 32-bit Cygwin, but wrong for 64-bit):
>>>
>>> long lrint (double x)
>>> {
>>>   long retval = 0L;
>>> #if defined(_AMD64_) || defined(__x86_64__) || defined(_X86_) ||
>>> defined(__i386__)
>>>   __asm__ __volatile__ ("fistpl %0"  : "=m" (retval) : "t" (x) :
>>> "st");

> Actually, I take it back.  gdb/objdump (and presumably binutils in
> general) was being deceptive to me about the nature of that mov
> instruction.  And in fact the fistpl should be fistpq.  That fixes it.

Is fistpq right on 32-bit, or is this a case where we need different assembly for 32-bit (fistpl) vs. 64-bit (fistql) to match the size of long that we are converting to?

--
Eric Blake, Principal Software Engineer
Red Hat, Inc.           +1-919-301-3266
Virtualization:  qemu.org | libvirt.org

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