What's the meaning of 'char (*p)[5];'?

Posted by jpmelos on Stack Overflow See other posts from Stack Overflow or by jpmelos
Published on 2011-06-24T12:12:55Z Indexed on 2011/06/24 16:22 UTC
Read the original article Hit count: 135

Filed under:
|
|

people.

I'm trying to grasp the differences between these three declarations:

char p[5];
char *p[5];
char (*p)[5];

I'm trying to find this out by doing some tests, because every guide of reading declarations and stuff like that has not helped me so far. I wrote this little program and it's not working (I've tried other kinds of use of the third declaration and I've ran out of options):

#include <stdio.h>                                                              
#include <string.h>                                                             
#include <stdlib.h>                                                             

int main(void) {                                                                
        char p1[5];                                                             
        char *p2[5];                                                            
        char (*p3)[5];                                                          

        strcpy(p1, "dead");                                                     

        p2[0] = (char *) malloc(5 * sizeof(char));                              
        strcpy(p2[0], "beef");                                                  

        p3[0] = (char *) malloc(5 * sizeof(char));                              
        strcpy(p3[0], "char");                                                  

        printf("p1 = %s\np2[0] = %s\np3[0] = %s\n", p1, p2[0], p3[0]);          

        return 0;                                                               
}

The first and second works alright, and I've understood what they do. What is the meaning of the third declaration and the correct way to use it?

Thank you!

© Stack Overflow or respective owner

Related posts about c

    Related posts about pointers