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Re: Should rand() return a RAND_MAX value for 32 bit target?
On 09/20/2018 11:05 AM, Vincent Lefevre wrote:
Actually, there may be a bug with -m32:
#include <stdio.h>
#include <float.h>
int main (void)
{
float f = 2147483646;
printf ("FLT_EVAL_METHOD = %d\n", (int) FLT_EVAL_METHOD);
if (f == 2147483647)
{
printf("%.20g\n", (double) f);
}
return 0;
}
$ gcc -std=c99 -m32 -O tst.c -o tst
$ ./tst
FLT_EVAL_METHOD = 2
2147483648
Since FLT_EVAL_METHOD = 2, float_t is long double, thus shouldn't
2147483647 be converted to long double directly (as f is regarded
as an operand of type long double), so that the result of the
equality test should be false?
The C standard says for FLT_EVAL_METHOD = 2: "evaluate all operations
and constants to the range and precision of the long double type".
^^^^^^^^^^^^^
Converting f first to float is contrary to the idea behind
FLT_EVAL_METHOD = 2, which is to avoid loss of precision in
intermediate operations.
Hrmmm ... I see FLT_EVAL_METHOD = 0 here with gcc 8.1.0 on sparc :
$ file ./flt_eval
./flt_eval: ELF 32-bit MSB executable SPARC32PLUS Version 1, V8+
Required, dynamically linked, not stripped, no debugging information
available
$ ./flt_eval
FLT_EVAL_METHOD = 0
2147483648
I see similar on ppc64 :
$ file flt_eval
flt_eval: ELF 32-bit MSB executable, PowerPC or cisco 4500, version 1
(SYSV), dynamically linked, interpreter /lib/ld.so.1, for GNU/Linux
3.2.0, not stripped
$ ./flt_eval
FLT_EVAL_METHOD = 0
2147483648
$
Not sure how you are getting FLT_EVAL_METHOD as 2.
What platform is this on ?
Dennis