类型转换整数指向c中的整数

Question

I saw somewhere in a pthreads program something like this...

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

void *fxn(void * t)
{
//some code
}

int main()
{
//some code
pthread_create(&thread,NULL,fxn,(int *)5);
//some code
}

Here, what is (int *)5 doing? and why is it not generating a warning/error? Also, how is it different from pthread_create(&thread,NULL,fxn,&x); where x=5 is of type int ?

Edit:

This works fine:

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

void * fxn(void *t)
{
    pthread_exit(t);
}
int main()
{
    pthread_t th;
    void *ret;
    int i=5;
    pthread_create(&th,NULL,fxn,&i);
    pthread_join(th,&ret);
    printf("%d\n",*(int *)ret);
}

but,

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

void * fxn(void *t)
{
    pthread_exit(t);
}
int main()
{
    pthread_t th;
    void *ret;
    int i=5;
    pthread_create(&th,NULL,fxn,(int *)5);
    pthread_join(th,&ret);
    printf("%d\n",*(int *)ret);
}

gives segmentation fault,core dumped...Why?

Answer 1

Technically, (int *)5 interprets 5 as an address. The result is typically a pointer to the address 0x0005 in your address space.

In this case, that's almost certainly not the real intention. (Such an address is almost always invalid if you're not doing low-level system stuff, at which point you probably don't have pthread_create available anyway.)

Here, since pthread_create takes a void * as some "extra data" to pass to the thread's startup function (in this case fxn ), whatever data you want to pass via that argument must be converted to a pointer.

The code in fxn will typically include an expression like (int)t to convert the pointer back into an integer. That's basically the only useful thing that can be done with it -- you're the only one that knows your pointer isn't really a pointer, and if anything (like, say, the last line of your main function...) ever tries to actually use it as a pointer, it's quite likely to trigger a segfault.

Answer 2

You don't get an error because by casting you tell the compiler that you know what you are doing.

The cast (int*)5 converts 5 to an int pointer pointing to some( very likely )invalid address. To get that value in the thread you have to cast void pointer back to an integer (int)t . Do not do this.

Passing &x , gets you a pointer to a variable x , which you can use in the thread function. Similar to this:

int x = 123 ;
int* px = &x ;
pthread_create(&thread,NULL,fxn,px);

And in the thread:

int* pt = t ;
printf( "%d" , *pt ) ;

Answer 3

(int*)5 means casting 5 to a pointer to an int.

This makes possible to pass a pointer which points to 5 in memory to the function pthread_create .

For example:

(int) 5; //cast to int
(int *) 5; //cast to pointer to int