Segmentation error when using function pointer

Question

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

The code is 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;
}

If I declare funptrinside main(), it gives me the correct conclusion. I would like to know what the problem is.

The problem is with the thread id. I used the same stream identifier "pid1" for both streams, and I tried to join "pid2", which also led to a segmentation error. Below is the straightened 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 . void *, , , (void *)funptr. void *(*) (void *), void*. . pthread_create

, SIGSEGV , pthread_create() pthread_t.

Answer 2

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

pid1.

pthread_join (2, NULL); pid2 - , ...