printf treats arguments of "%c" not as expected

[Yunfeng Wang]

Configuration Information [Automatically generated, do not change]:
Machine: i486
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='i486'
-DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='i486-pc-linux-gnu'
-DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash'
-DSHELL -DHAVE_CONFIG_H   -I.  -I../bash -I../bash/include
-I../bash/lib   -g -O2 -Wall
uname output: Linux loco 2.6.38-2-686 #1 SMP Tue Mar 29 17:27:45 UTC
2011 i686 GNU/Linux
Machine Type: i486-pc-linux-gnu

Bash Version: 4.1
Patch Level: 5
Release Status: release

Description:
    The builtin printf does not print arguments of format "%c" as described
    in printf(3). /usr/bin/printf from coreutils has the same problem.

Repeat-By:

    $ printf %c 65 66 67
    666

    The expected output is ABC, i.e. characters with ASCII code of 65 66 67

Fix:

--- bash-4.1/builtins/printf.def        2009-11-21 04:31:23.000000000 +0800
+++ bash-4.1-new/builtins/printf.def    2011-06-22 19:25:30.000000000 +0800
@@ -385,9 +385,9 @@
            {
            case 'c':
              {
-               char p;
+               int p;

-               p = getchr ();
+               p = getintmax ();
                PF(start, p);
                break;
              }