vsprintf: correctly handle width when '#' flag used in %#p format

The '%p' output of the kernel's vsprintf() uses spec.field_width to
determine how many digits to output based on 2 * sizeof(void*) so that all
digits of a pointer are shown.  ie.  a pointer will be output as
"001A2B3C" instead of "1A2B3C".  However, if the '#' flag is used in the
format (%#p), then the code doesn't take into account the width of the
'0x' prefix and will end up outputing "0x1A2B3C" instead of "0x001A2B3C".

This patch reworks the "pointer()" format hook to include 2 characters for
the '0x' prefix if the '#' flag is included.

[akpm@linux-foundation.org: checkpatch fixes]
Signed-off-by: Grant Likely <grant.likely@secretlab.ca>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
This commit is contained in:
Grant Likely 2012-05-31 16:26:08 -07:00 committed by Linus Torvalds
parent d84970bbaf
commit 725fe002d3

View File

@ -870,13 +870,15 @@ static noinline_for_stack
char *pointer(const char *fmt, char *buf, char *end, void *ptr,
struct printf_spec spec)
{
int default_width = 2 * sizeof(void *) + (spec.flags & SPECIAL ? 2 : 0);
if (!ptr && *fmt != 'K') {
/*
* Print (null) with the same width as a pointer so it makes
* tabular output look nice.
*/
if (spec.field_width == -1)
spec.field_width = 2 * sizeof(void *);
spec.field_width = default_width;
return string(buf, end, "(null)", spec);
}
@ -931,7 +933,7 @@ char *pointer(const char *fmt, char *buf, char *end, void *ptr,
*/
if (in_irq() || in_serving_softirq() || in_nmi()) {
if (spec.field_width == -1)
spec.field_width = 2 * sizeof(void *);
spec.field_width = default_width;
return string(buf, end, "pK-error", spec);
}
if (!((kptr_restrict == 0) ||
@ -948,7 +950,7 @@ char *pointer(const char *fmt, char *buf, char *end, void *ptr,
}
spec.flags |= SMALL;
if (spec.field_width == -1) {
spec.field_width = 2 * sizeof(void *);
spec.field_width = default_width;
spec.flags |= ZEROPAD;
}
spec.base = 16;