Using the %s didn't solve the problem:
erdav...@antares ~/perl/feedbacks
$ cat BUG.c
#include <stdio.h>
int main() {
const char * str =
"0123456789" // 0 - 9
"0123456789" // 10 - 19
"0123456789" // 20 - 29
"0123456789" // 30 - 39
"0123456789" // 40 - 49
"0123456789" // 50 - 59
"0123456789" // 60 - 69
"0123456789" // 70 - 79
"0123456789" // 80 - 89
"0123456789" // 90 - 99
"0123456789" // 100 - 109
"0123456789" // 110 - 119
"0123456ç89" // 120 - 127
;
printf("%s", str);
return 0;
}
erdav...@antares ~/perl/feedbacks
$ gcc -Wall BUG.c -o BUG
erdav...@antares ~/perl/feedbacks
$ ./BUG
ç89
Eduardo R. D'Avila
2010/1/4 Eric Blake <e...@byu.net>:
> Eduardo D'Avila <erdavila <at> gmail.com> writes:
>
>> I've found a bug that happens when the 128th (index 127 on a 0-based
>> string) char of a string is a multibyte char. When I print such
>> string, only the multibyte char and the chars after it are displayed.
>
> What you've found is a bug in your own program, at lesat for the BUG.c version
> of your report.
>
> http://cygwin.com/ml/cygwin/2010-01/msg00100.html
>
> Try 'printf ("%s",str)' rather than 'printf (str)' to see the difference. And
> why 128 bytes into the string? That's the cutoff of where gcc optimizes a
> printf without % into a puts.
>
> --
> Eric Blake
> --
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