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arrayscpointersinitializationdeclaration
bursikif
bursikif

Reputation: 21

Creating an array whose elements are pointers to different arrays

I'm trying to create an array of pointers. In this code, array 'p' is supposed to contain pointers to the other arrays 'a', 'b', and 'c'. I can't figure out what is wrong with my code. Any help?

#include <stdio.h>
int main() {
    int a[3]={'4','1','3'};
    int b[3]={'a','1','3'};
    int c[3]={'y','1','3'};
    int *p[2];
    
    p[1]=a[1];
    *(p+1)= a+1;
    printf("%d",p[1]);
    return 0;
}

Upvotes: 1

Views: 60

Answers (1)

Vlad from Moscow
Vlad from Moscow

Reputation: 310950

To declare an array of pointers to these arrays

int a[3]={'4','1','3'};
int b[3]={'a','1','3'};
int c[3]={'y','1','3'};

you need to write

int * p[3] = { a, b, c };

In this declaration the array designators used as initializers are implicitly converted to pointers to their first elements. That is it is the same if to write

int * p[3] = { &a[0], &b[0], &c[0] };

This expression statement

p[1]=a[1];

is incorrect because the left side operand has the type int * while the right side operand has the type int.

This statement

*(p+1)= a+1;

that is equivalent to the statement

p[1] = a + 1;

or to

p[1] = &a[1];

is correct.

In this statement

printf("%d",p[1]);

there is used an incorrect conversion specifier %d with pointer expression p[1].

If you want to output the pointer expression p[1] then you need to write

printf( "%p\n", ( void * )p[1] );

If you want to output the pointed value by the expression p[1] you need to write

printf( "%d\n", *p[1] );

Or if you want to output it as a character you can write

printf( "%c\n", *p[1] );

Upvotes: 4

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