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[Bug c/35774] New: wrong code on casting int result to signed char
- From: "hongbo dot yang at mathworks dot com" <gcc-bugzilla at gcc dot gnu dot org>
- To: gcc-bugs at gcc dot gnu dot org
- Date: 31 Mar 2008 14:33:22 -0000
- Subject: [Bug c/35774] New: wrong code on casting int result to signed char
- Reply-to: gcc-bugzilla at gcc dot gnu dot org
In the main function there is a cast casting int result to signed char.
typedef int int32_T;
typedef unsigned int uint32_T;
typedef signed char int8_T;
#include <stdio.h>
int main(void)
{
int32_T i;
uint32_T numerator;
for (i = 126; i < 256; i++) {
numerator = (uint32_T)((int8_T)(-128 + i) + 128) * 63U;
fprintf(stdout, "i = %d numerator = %u\n", i, numerator);
}
return numerator;
}
This program is compiled with gcc -O0 -g,
I suppose the result of the variable "numerator" should be:
126 * 63
127 * 63
128 * 63
....
But the printed result is equal to:
126 * 63
127 * 63
- 128 * 63
....
The generated assembly code on mac with Intel Core-2-Duo processor is:
movl -24(%ebp), %eax
movsbl %al,%edx
movl %edx, %eax
sall $6, %eax
subl %edx, %eax
movl %eax, -20(%ebp)
It seems both -128 and +128 are ignored.
This problem doesn't appear on gcc 4.1.1.
--
Summary: wrong code on casting int result to signed char
Product: gcc
Version: 4.0.1
Status: UNCONFIRMED
Severity: major
Priority: P3
Component: c
AssignedTo: unassigned at gcc dot gnu dot org
ReportedBy: hongbo dot yang at mathworks dot com
GCC build triplet: i686-linux-gnu
GCC host triplet: i686-linux-gnu
GCC target triplet: i686-linux-gnu
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=35774