why gdb show wrong variable value?
c, gdb
Solution
This is a bug that is fixed if you compile with `-fvar-tracking`. Your question is a tighter version of this SO question, which references a bug report on GCC 4.8.0 suggesting the above compile flag.
Problem
I have simple program: ``` #include <stdio.h> void func(int i) { 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 at 0x400543: file test.c, line 9. (gdb) run Starting program: /tmp/test Breakpoint 1, main (argc=1, argv=0x7fffffffe458) at test.c:9 9 func(0); (gdb) s func (i=0) at test.c:4 4 i =1; (gdb) p i $1 = 0 (gdb) n 5 printf("%d\n", i); (gdb) p i $2 = 0 (gdb) ``` Program works fine, shows "1", but why gdb shows me "0" value? Debian wheezy. I observed that on gcc-4.7, gcc-4.6. On gcc-4.4 all is ok.