[SOLVED] Passing array to thread function

zak100 · 08-12-2010, 12:12 AM

Hi,
I am trying to pass an array to thread program. Its printing the val[0] value but its not printing value at val[1]. Can somebody help me with this problem.

Code:

  

#include <unistd.h>
#include <stdlib.h>
#include <pthread.h>
#include <stdio.h>

unsigned int counter=0;
pthread_mutex_t mutex= PTHREAD_MUTEX_INITIALIZER;

static void *thread_func(void *arg){
int val[2];
int i;

val[0]=*(int *)arg;
val[1]=*(int *)(arg+1);
printf("val[0]=%d      val[1]=%d", val[0], val[1]);
} 

main(){
pthread_t thread1, thread2;
int thread1_stuff[2];

thread1_stuff[0]=110;
thread1_stuff[1]=10;

pthread_create(&thread1, NULL, thread_func, thread1_stuff);
//pthread_create(&thread2, NULL, thread_func, &thread2_stuff);
pthread_join(thread1, NULL);
//pthread_join(thread2, NULL);
return EXIT_SUCCESS;
}

Zulfi.

paulsm4 · 08-12-2010, 12:42 AM

Hi -

Suggested change:

Code:

static void *thread_func(void *arg){
  int *val_p = (int *) arg;

  fprintf(stderr, "val[0]=%d, val[1]=%d\n", val_p[0], val_p[1]);
  ...

zak100 · 08-12-2010, 01:27 AM

I have tried this. Its working. Thanks for your help.

Zulfi.

Wim Sturkenboom · 08-12-2010, 01:51 AM

When I compiled and ran the code, val[1] had a value, but it was the wrong one.

Rewritten your thread function a bit to (hopefully) show where it goes wrong

Code:

static void *thread_func(void *arg){
    int val[2];
    int i;

    val[0]=*(int *)arg;
    val[1]=*(int *)(arg+1);
    printf("val[0]=%d (0x%08X)     val[1]=%d (0x%08X)\n", val[0], val[0], val[1], val[1]);

    for(i=0;i<sizeof(int)*2;i++)
    {
        printf("%p -> 0x%02X\n", arg+i, *(unsigned char*)(arg+i));
    }
    printf("\n");

    return NULL;
}

Output:

Code:

wim@btd-techweb01:~/forums_linuxquestions/lq825699$ ./lq825699
val[0]=110 (0x0000006E)     val[1]=167772160 (0x0A000000)
0xbffff798 -> 0x6E }arg
0xbffff799 -> 0x00 }    }arg+1
0xbffff79a -> 0x00 }    }
0xbffff79b -> 0x00 }    }
0xbffff79c -> 0x0A      }
0xbffff79d -> 0x00
0xbffff79e -> 0x00
0xbffff79f -> 0x00

It displays the address locations of arg (for two integers) and the values that are stored in there.

paulsm4 · 08-12-2010, 11:07 AM

Wim -

You're absolutely right, but let me spell it out:

Code:

  /* Intent: cast "arg" to an int pointer, get the second int value */
  val[1]=*(int *)(arg+1);

  /* Actuality:
     "arg" and "arg+1" are both BYTE pointers (per your sample output)
     We cast "arg+1" to an int pointer...
     ... but we're already screwed...
     ... because we already have the wrong address
   */

Code:

  /* Better (possibly: I haven't actually tried it ;)) */
  val[1] = *((int *)arg)+1)

Code:

  /* Best: just cast the pointer itself, up-front, then just use it */
  int *val_p = (int *) arg;
  val[0] = val_p[0];
  val[1] = val_p[1];