Invalid output when both the 0 flag and a precision appear

[GabrielRavier]

With the following code:

#include <stdio.h>

#ifdef USE_NANOPRINTF
#define NANOPRINTF_IMPLEMENTATION
#define NANOPRINTF_USE_FIELD_WIDTH_FORMAT_SPECIFIERS 1
#define NANOPRINTF_USE_PRECISION_FORMAT_SPECIFIERS 1
#define NANOPRINTF_USE_FLOAT_FORMAT_SPECIFIERS 1
#define NANOPRINTF_USE_LARGE_FORMAT_SPECIFIERS 1
#define NANOPRINTF_USE_BINARY_FORMAT_SPECIFIERS 1
#define NANOPRINTF_USE_WRITEBACK_FORMAT_SPECIFIERS 1
#include "nanoprintf.h"
#define snprintf npf_snprintf
#endif

int main()
{
    char output[100];
    snprintf(output, sizeof(output), "%02.1d", 0);
    printf("%s\n", output);
}

when compiled without USE_NANOPRINTF, I get an output of 0, whereas when using nanoprintf, 00 is output:

$ gcc test.c -o default
$ gcc -DUSE_NANOPRINTF test.c -o nanoprintf
$ ./default; ./nanoprintf 
 0
00

Nanoprintf's behavior is invalid because the C standard states that:

For d, i, o, u, x, and X conversions, if a precision is specified, the 0 flag is ignored.

The current trunk version of nanoprintf fails to do so.

[charlesnicholson]

Thanks for the report! I'll investigate this ASAP, sorry!

[charlesnicholson]

With #245 and your test driver, I get:

❯ gcc test.c -o default
❯ gcc -DUSE_NANOPRINTF test.c -o npf_test
❯ ./default;./npf_test
 0
 0

I'll cut a new release for this after I merge it.

[charlesnicholson]

Released v0.3.4 with this fix.
https://github.com/charlesnicholson/nanoprintf/releases/tag/v0.3.4

Thanks again! I even had it as a requirement in the implementation comments, but didn't adhere to it. Sigh.