CODEWITHSUNDEEP
cstaticgdballocation
Debashish
Debashish

Reputation: 1185

When static variables are created in c language

Here gdb does not stop at Line:4.

enter image description here

Next,

Without hitting the declaration line at Line:5, variable x is existing and initialized.

enter image description here

Next,

But here it shows out of scope (yes it should according to me).

enter image description here

Now, I have the following doubts regarding this particular instance of c program.

  1. When exactly the memory for variable x in P1() gets created and initialized?
  2. why gdb did not stop at static declaration statement in inside P1() in the first example?
  3. If we call P1() again will the program control simply skip the declaration statement?

Upvotes: 2

Views: 691

Answers (1)

dbrank0
dbrank0

Reputation: 9476

It has already been explained (in related topics linked in comments below question) how static variables work.

Here is actual code generated by a gcc for your p1 function (by gcc -c -O0 -fomit-frame-pointer -g3 staticvar.c -o staticvar.o) then disassembled with related source.

Disassembly of section .text:

0000000000000000 <p1>:
#include <stdio.h>

void p1(void)
{
   0:   48 83 ec 08             sub    $0x8,%rsp
    static int x = 10;
    x += 5;
   4:   8b 05 00 00 00 00       mov    0x0(%rip),%eax        # a <p1+0xa>
   a:   83 c0 05                add    $0x5,%eax
   d:   89 05 00 00 00 00       mov    %eax,0x0(%rip)        # 13 <p1+0x13>
    printf("%d\n", x);
  13:   8b 05 00 00 00 00       mov    0x0(%rip),%eax        # 19 <p1+0x19>
  19:   89 c6                   mov    %eax,%esi
  1b:   bf 00 00 00 00          mov    $0x0,%edi
  20:   b8 00 00 00 00          mov    $0x0,%eax
  25:   e8 00 00 00 00          callq  2a <p1+0x2a>
}
  2a:   90                      nop
  2b:   48 83 c4 08             add    $0x8,%rsp
  2f:   c3                      retq

So, as you see there is no code for declaration of x. GDB can only break on actual machine code instruction and as there is none, it breaks on next instruction (mov), which matches line 5.

Upvotes: 1

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