[Bug stdio/21360] New: snprintf %n does not conform to ISO C when characters are not printed

[vincent-srcware at vinc17 dot net]

https://sourceware.org/bugzilla/show_bug.cgi?id=21360

            Bug ID: 21360
           Summary: snprintf %n does not conform to ISO C when characters
                    are not printed
           Product: glibc
           Version: 2.24
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: stdio
          Assignee: unassigned at sourceware dot org
          Reporter: vincent-srcware at vinc17 dot net
  Target Milestone: ---

Consider the following example:

#include <stdio.h>

int main(void)
{
  int r, n = -1;

  r = snprintf (NULL, 0, "foo%n", &n);
  printf ("%d %d\n", r, n);
  return 0;
}

With glibc 2.24, I get 3 3 instead of the expected 3 0. Indeed, the glibc
manual (whose description seems equivalent to ISO C11) says for %n:

12.12.6 Other Output Conversions
--------------------------------
[...]
   The '%n' conversion is unlike any of the other output conversions.
It uses an argument which must be a pointer to an 'int', but instead of
printing anything it stores the number of characters printed so far by
this call at that location.

but due to the size 0, nothing has been printed. Thus one should have n = 0.

Note that in its snprintf description:

     The return value is the number of characters which would be
     generated for the given input, excluding the trailing null.

"would be", thus the expected r = 3. There is no such "would be" for %n. In the
C11 draft I have, this is said in a similar way:

     The snprintf function returns the number of characters that would
     have been written had n been sufficiently large, [...]

The "had n been sufficiently large" is quite explicit. But again, there is no
such thing for %n.

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