c - why gdb show wrong variable value? -

this question has answer here:

• gdb prints wrong values when modifying arguments 3 answers

i have simple program:

#include <stdio.h>  void func(int i) {     = 1;     printf("%d\n", i); }  int main(int argc, char *argv[]){     func(0);     return 0; } 

and now:

gcc test.c -g -o test  gdb test (gdb) b main breakpoint 1 @ 0x400543: file test.c, line 9. (gdb) run starting program: /tmp/test   breakpoint 1, main (argc=1, argv=0x7fffffffe458) @ test.c:9 9       func(0); (gdb) s func (i=0) @ test.c:4 4       =1; (gdb) p $1 = 0 (gdb) n 5       printf("%d\n", i); (gdb) p $2 = 0 (gdb) 

program works fine, shows "1", why gdb shows me "0" value?

debian wheezy.

i observed on gcc-4.7, gcc-4.6. on gcc-4.4 ok.

this bug fixed if compile -fvar-tracking. question tighter version of this question, references bug report on gcc 4.8.0 suggesting above compile flag.