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Date:	Fri, 6 Dec 2024 21:26:55 -0800
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Subject:	Re: Bug in handling of "1$", "2$" in printf format
To:	The Cygwin Mailing List <cygwin AT cygwin DOT com>
Cc:	Keith Thompson <Keith DOT S DOT Thompson AT gmail DOT com>
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From:	Keith Thompson via Cygwin <cygwin AT cygwin DOT com>
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Brian Inglis wrote:
> On 2024-12-06 19:16, Keith Thompson via Cygwin wrote:
> > The use of "1$", "2$" et al in printf format specifiers is a
> > POSIX-specific feature.
> >
> > On Cygwin (newlib) this is handled correctly in most cases, but one
> > example I tried misbehaves.
> > The output is correct on other implementations, including glibc and
> > musl on Ubuntu.
> >
> > This C program:
> >
> > #include <stdio.h>
> > int main(void) {
> >      long long a = 123456789876543210;
> >      double b=1.0/3;
> >      printf("a:%2$8.8lld b:%1$10.2g\n", b, a);
> > }
> >
> > should produce this output:
> >
> > a:123456789876543210 b:      0.33
> >
> > Under Cygwin (fully updated), with "gcc c.c -o c && ./c", the output is:
> >
> > a:140732550844138 b:  7.1e-315
> >
>
> Confirmed with gcc 12.4 and minor tweaks to constant data types: printf is
> ignoring arg positions:
[SNIP]

It's not always ignoring arg positions. I think there's an interaction
between the "1$" / "2$" position specification and relatively complex
format specifiers.  The following case works correctly:

$ cat c2.c
#include <stdio.h>
int main(void) {
    int a = 42;
    double b = 1.0/3.0;
    printf("a:%2$d b:%1$g\n", b, a);
    printf("a:%1$d b:%2$g\n", a, b);
    printf("a:%d b:%g\n",     a, b);
}
$ gcc c2.c -o c2 && ./c2
a:42 b:0.333333
a:42 b:0.333333
a:42 b:0.333333
$

(And the version of gcc shouldn't matter.  printf is implemented in
newlib.  The code in question should be in newlib/libc/stdio/vfprintf.c.)

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