Package: libc6
Version: 2.3.6.ds1-13etch2
I found that strtoul doesn't accept "i" as a digit of a
base 36 number under tr_TR locale.
% cat tst.c
#include <stdio.h>
#include <locale.h>
#include <stdlib.h>
#include <errno.h>
int main(int argc, char **argv)
{
unsigned long ret;
char *e = NULL;
setlocale(LC_ALL, "");
errno = 0;
ret = strtoul("i", &e, 36);
if (errno != 0) { perror("strtoul"); }
printf("%lu [%s]\n", ret, e);
return 0;
}
% gcc -Wall tst.c
% LANG=tr_TR ./a.out
0 [i]
It returns 0. "i" is not converted.
It returns 18 under C locale.
% LANG=C ./a.out
18 []
It is caused by Turkish case conversion rule.
strtoul(3) describes about locale as follows:
| In locales other than the "C" locale, other strings may be accepted.
| (For example, the thousands separator of the current locale may be sup-
| ported.)
I think it is not intentional to reject "i" under tr_TR
locale.
% dpkg -l|grep libc6
ii libc6 2.3.6.ds1-13etch2
GNU C Library: Shared libraries
ii libc6-dev 2.3.6.ds1-13etch2
GNU C Library: Development Libraries and Hea
ii libc6-i686 2.3.6.ds1-13etch2
GNU C Library: Shared libraries [i686 optimi
--
Tanaka Akira
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