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usage.c: detect recursion in die routines and bail out immediately 

It is theoretically possible for a die handler to get into a state of
infinite recursion.  For example, if a die handler called another function
which itself called die().  Let's at least detect this situation, inform the
user, and call exit.

Signed-off-by: Brandon Casey <bcasey@nvidia.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
maint
Brandon Casey 2012-11-14 17:45:52 -08:00 committed by Junio C Hamano
parent 7e2010537e
commit cd163d4b4e
1 changed files with 15 additions and 0 deletions
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15
usage.c
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@ -6,6 +6,8 @@
#include "git-compat-util.h" #include "git-compat-util.h"
#include "cache.h" #include "cache.h"


static int dying;

void vreportf(const char *prefix, const char *err, va_list params) void vreportf(const char *prefix, const char *err, va_list params)
{ {
char msg[4096]; char msg[4096];
@ -82,6 +84,12 @@ void NORETURN die(const char *err, ...)
{ {
va_list params; va_list params;


if (dying) {
fputs("fatal: recursion detected in die handler\n", stderr);
exit(128);
}
dying = 1;

va_start(params, err); va_start(params, err);
die_routine(err, params); die_routine(err, params);
va_end(params); va_end(params);
@ -94,6 +102,13 @@ void NORETURN die_errno(const char *fmt, ...)
char str_error[256], *err; char str_error[256], *err;
int i, j; int i, j;


if (dying) {
fputs("fatal: recursion detected in die_errno handler\n",
stderr);
exit(128);
}
dying = 1;

err = strerror(errno); err = strerror(errno);
for (i = j = 0; err[i] && j < sizeof(str_error) - 1; ) { for (i = j = 0; err[i] && j < sizeof(str_error) - 1; ) {
if ((str_error[j++] = err[i++]) != '%') if ((str_error[j++] = err[i++]) != '%')