Segmentation fault while using function pointer

Question

I get a segmentation fault when I declare a function pointer before main() and assign it with the address of a function inside main. What is the actual problem that occurs if the function pointer is declared before main()??

The code is given below:

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

void fun1(char *str)
{
    printf("%s",str);
}

void (* funptr)(char *);

int main()
{

    char msg1[10]="Hi";
    char msg2[10]="Hello";
    pthread_t pid1, pid2;

    funptr=&fun1;

    pthread_create(&pid1,NULL,(void *)(*funptr),(void *)msg1);
    pthread_create(&pid1,NULL,(void *)(*funptr),(void *)msg2);

    pthread_join(pid1,NULL);
    pthread_join(pid2,NULL);
    return 0;
}

Whereas when I declare funptr inside main(), it gives me the proper output. Would like to know what exactly is the issue.

The problem was with the thread id. I had used the same thread id "pid1" for both the threads and I was trying to join "pid2" also which resulted in segmentation fault. Following is the rectified code...

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

void fun1(char *str)
{
    printf("%s",str);
}

 void (* funptr)(char *);

int main()
{

    char msg1[10]="Hi";
    char msg2[10]="Hello";
    pthread_t pid1, pid2;

    funptr=&fun1;

    pthread_create(&pid1,NULL,(void *)(*funptr),(void *)msg1);
    pthread_create(&pid2,NULL,(void *)(*funptr),(void *)msg2);
    pthread_join(pid1,NULL);
    pthread_join(pid2,NULL);
    return 0;
}

Answer 1

funptr is already a function pointer. To cast it to void *, all you need is (void *)funptr. You are required to have the third argument of type void *(*) (void *), not cast your function to void*. See pthread_create documentation

As Santhosh has written in the comments, the reason for the SIGSEGV is that pthread_create() is given as argument pointer to the same pthread_t.