Re: A simple query about memory mgmt

["Peter Teoh" <htmldeveloper@xxxxxxxxx>]

On Wed, Aug 13, 2008 at 1:28 AM, Santosh Pradhan
<santosh.pradhan@xxxxxxxxx> wrote:
> Hi All,
> I am new to this group and I have a question regarding following C program.
>
> #include<stdio.h>
>
> #define NAME "santosh"
>
> int main()
> {
>     char *p_name = NAME;
>     char *q_name = NAME;
>     if (p_name == q_name)
>          printf("Hello, World\n");
>     return 0;
> }
>

p_name and q_name are variable allocated from the stack, and it will
always be different address from the stack.   Yes, the optimizer can
shrink them to the same address, but whether the optimizer does it or
it does not matter.   Because u are not comparing the address - which
is &p_name == &q_name, but u are comparing the value inside the
addresses.

But since u have assigned it to the same address of NAME, it will
always print HELLO world.   So the whole thing has nothing got to do
with optimization (gcc -O0 to disable it, which is also default).

First, modifying your program to:

#include<stdio.h>

#define NAME "santosh"

int main()
{
    char *p_name = NAME;
    char *q_name = NAME;

    printf("ptr addr %p %p\n", &p_name, &q_name);
    printf("ptr addr %p %p\n", p_name, q_name);
    if (p_name == q_name)
         printf("Hello, World\n");
    return 0;
}

and then compile and running it under GDB and displaying the assembly:

(gdb) break main
Breakpoint 1 at 0x80483e5: file santosh.c, line 8.
(gdb) run
Starting program: /root/a.out

Breakpoint 1, main () at santosh.c:8
8           char *p_name = NAME;
(gdb) disassemble  $eip
Dump of assembler code for function main:
0x080483d4 <main+0>:    lea    ecx,[esp+4]
0x080483d8 <main+4>:    and    esp,0xfffffff0
0x080483db <main+7>:    push   DWORD PTR [ecx-4]
0x080483de <main+10>:   push   ebp
0x080483df <main+11>:   mov    ebp,esp
0x080483e1 <main+13>:   push   ecx
0x080483e2 <main+14>:   sub    esp,0x24
0x080483e5 <main+17>:   mov    DWORD PTR [ebp-8],0x8048520=======>ptr for p_name
0x080483ec <main+24>:   mov    DWORD PTR
[ebp-12],0x8048520========>ptr for q_name
0x080483f3 <main+31>:   lea    eax,[ebp-12]
0x080483f6 <main+34>:   mov    DWORD PTR [esp+8],eax
0x080483fa <main+38>:   lea    eax,[ebp-8]
0x080483fd <main+41>:   mov    DWORD PTR [esp+4],eax
0x08048401 <main+45>:   mov    DWORD PTR [esp],0x8048528
0x08048408 <main+52>:   call   0x80482d8 <printf@plt>
0x0804840d <main+57>:   mov    eax,DWORD PTR [ebp-12]
0x08048410 <main+60>:   mov    edx,DWORD PTR [ebp-8]
0x08048413 <main+63>:   mov    DWORD PTR [esp+8],eax
0x08048417 <main+67>:   mov    DWORD PTR [esp+4],edx
0x0804841b <main+71>:   mov    DWORD PTR [esp],0x8048528
0x08048422 <main+78>:   call   0x80482d8 <printf@plt>
0x08048427 <main+83>:   mov    edx,DWORD PTR [ebp-8]
0x0804842a <main+86>:   mov    eax,DWORD PTR [ebp-12]
0x0804842d <main+89>:   cmp    edx,eax==================> u are
comparing the ptr value, not the addresses of the ptr, but u have just
set the ptr to be the same earlier?   so it will always be the same
0x0804842f <main+91>:   jne    0x804843d <main+105>
0x08048431 <main+93>:   mov    DWORD PTR [esp],0x8048538
0x08048438 <main+100>:  call   0x80482e8 <puts@plt>
0x0804843d <main+105>:  mov    eax,0x0
0x08048442 <main+110>:  add    esp,0x24
0x08048445 <main+113>:  pop    ecx
0x08048446 <main+114>:  pop    ebp
0x08048447 <main+115>:  lea    esp,[ecx-4]
0x0804844a <main+118>:  ret
End of assembler dump.
(gdb)

Not sure if I had fully understood your problem?

-- 
Regards,
Peter Teoh

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