Hi Randi,
You are observing this because of malloc's behaviour. If you allocate
and free same-sized memory multiple times, then malloc tries to reuse
the memory. You can try this by putting malloc/free statements in a
loop.
In your case, first thread gets created, it starts executing, it
allocates the memory, prints the address and frees it. After first
thread finishes its execution, second thread starts. It allocates the
memory so allocator reuses the same chunk and you get same address.
If you put some sleep between malloc and free OR remove the free
statement itself as you did then you will get different addresses. So
this is really a timing issue when you do thread programming.
Hope this helps you understand the issue.
-Vinit
On Fri, Dec 25, 2009 at 12:11 AM, Randi Botse <nightdecoder@xxxxxxxxx> wrote:
> Hi all,
>
> I'm now learning thread reentrancy on Linux, so i write simple codes
> for demonstration:
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <pthread.h>
>
> void *func(void *data)
> {
> char *ptr;
>
> ptr = malloc(8);
> printf("thread%i -> ptr at: %p\n", *(int*) data, ptr);
> free(ptr);
> return NULL;
> }
>
> #define NTHREAD 5
>
> int main(void)
> {
> pthread_t thread[NTHREAD];
> int i;
>
> for (i = 0; i < NTHREAD; i++)
> if (pthread_create(&thread[i], NULL, func, (void*) &i))
> return -1;
> for (i = 0; i < NTHREAD; i++)
> pthread_join(thread[i], NULL);
> return 0;
> }
>
>
> I know the func() function is not reentrant, worst printf() debugging
> told me that's each thread pointer to character (ptr) may has same
> address, like this output:
>
> thread1 -> ptr at: 0x9a80128
> thread1 -> ptr at: 0x9a80138
> thread2 -> ptr at: 0x9a801d8
> thread4 -> ptr at: 0x9a801d8
> thread4 -> ptr at: 0x9a801d8
>
> I try to modify the func() by not free ptr to be:
>
> void *func(void *data)
> {
> char *ptr;
>
> ptr = malloc(8);
> printf("thread%i -> ptr at: %p\n", *(int*) data, ptr);
> /* free(ptr); */
> return NULL;
> }
>
> And the result is i always got different address of each thread's ptr, such as:
>
> thread0 -> ptr at: 0x8db2098
> thread1 -> ptr at: 0x8db2138
> thread2 -> ptr at: 0x8db21d8
> thread3 -> ptr at: 0x8db2278
> thread4 -> ptr at: 0x8db2318
>
> Can you explain me why this happen? how about my printf() debugging
> method? it's works for this demonstration?
> Thanks before.
>
> - Randi
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