CODEWITHSUNDEEP
carrays
mysql lover
mysql lover

Reputation: 59

Print the memory location of an array

Im new to the C programming language. I have a pointer array with 2 elements. I want to print the values inside the array. The code I have written

#include<stdio.h>
int main ()
{
int *bae[] = {10,12};
int i;
for(i=0;i<2;i++) {
printf("%d",*bae[i]);
}
return 0;
}

When I run this code it gave me warning like:

warning: (near initialization for 'bae[0]') [enabled by default]
warning: initialization makes pointer from integer without a cast [enabled by default]
warning: (near initialization for 'bae[1]') [enabled by default]

Hope you could help me in finding the solution

Upvotes: 0

Views: 300

Answers (3)

Karthick
Karthick

Reputation: 1000

int *bae[] = {10,12};

It means bae is an array of integer pointers.

printf("%d",*bae[i]);

The above statement is equal to *(bae + i). so it will try to access the value at the address 10 and 11 which leads to undefined behavior.

Upvotes: 0

Jashaszun
Jashaszun

Reputation: 9270

bae is an array of int-pointers at the moment. If you want just an array, it should be int bae[] = {10, 12};

Then, you can printf("%d", bae[i]);

What the warning is saying ("initialization makes pointer from integer without a cast") is that you are converting 10 and 12 into int* when storing them in bae.

EDIT: If you really want to use a pointer array, then you could do something like this (though not advisable, and in no way better than my previous answer):

#include<stdio.h>
int main ()
{
    int ten = 10;
    int twelve = 12;
    int *bae[] = { &ten, &twelve };
    int i;
    for(i=0;i<2;i++) {
        printf("%d",*bae[i]);
    }
    return 0;
}

This would now print 10 and 12.

Upvotes: 2

Eugeniu Rosca
Eugeniu Rosca

Reputation: 5305

#include <stdio.h>

int main ()
{
    int bae[] = {10,12};
    int i;
    for(i=0; i<2; i++) {
        printf("%d\n", bae[i]);
    }
    return 0;
}

produces the output:

10
12

Upvotes: 2

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