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Re: serious sscanf bug
From: |
Will Styles |
Subject: |
Re: serious sscanf bug |
Date: |
Wed, 30 Apr 2003 04:02:12 -0700 (PDT) |
--- Andreas Schwab <address@hidden> wrote:
> Will Styles <address@hidden> writes:
>
> |> program:
> |> #include <stdio.h>
> |>
> |> int main (void)
> |> {
> |> int val;
> |> sscanf ("05", "%02i", &val);
> |> printf ("val = %i\n", val);
> |>
> |> sscanf ("06", "%02i", &val);
> |> printf ("val = %i\n", val);
> |>
> |> sscanf ("07", "%02i", &val);
> |> printf ("val = %i\n", val);
> |>
> |> sscanf ("08", "%02i", &val);
> |> printf ("val = %i\n", val);
> |>
> |> sscanf ("09", "%02i", &val);
> |> printf ("val = %i\n", val);
> |> }
> |>
> |> output:
> |> val = 5
> |> val = 6
> |> val = 7
> |> val = 0
> |> val = 0
>
> Correct.
>
> |> what i should get:
> |> val = 5
> |> val = 6
> |> val = 7
> |> val = 8
> |> val = 9
>
> Wrong. Look up %i in the manual.
>
ok, i feel stupid now - trying to read in octals :)
i always thought that %i == %d.
so this bug should be closed (if it isn't already).
i'm just wondering though: should it be "%02d" or
"%2d" or "%d" to read in an integer padded with up to
2 zeros or is that printf syntax only?
thanks
- will
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