What's the output?

<syntaxhighlight lang="c" name="pointer_1"> int main() {

       int arr[3] = {2, 3, 4};
       char *p;
       p = (char*)arr;
       printf("%d ", *p);
       p = p+1;
       printf("%d\n", *p);
       
       return 0;

}


</syntaxhighlight>


(A) 2 3

(B) 2 0

(C) 1 0

(D) Garbage value

Solution by Arjun Suresh

Assume start location of arr s 1000. So, the first two elements of the array will be stored in memory as follows:

1000: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes)
1004: 0 0 0 0 0 0 0 3

Now, %d, *p will print the first 4 bytes starting at 1000, that is 2. When p is incremented, it will go to 1001, as p is a char pointer.

1001: 0 0 0 0 0 0 0 0

So, output will be 0.





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What's the output?[edit]

<syntaxhighlight lang="c" name="pointer_1"> int main() {

       int arr[3] = {2, 3, 4};
       char *p;
       p = (char*)arr;
       printf("%d ", *p);
       p = p+1;
       printf("%d\n", *p);
       
       return 0;

}


</syntaxhighlight>


(A) 2 3

(B) 2 0

(C) 1 0

(D) Garbage value

Solution by Arjun Suresh[edit]

Assume start location of arr s 1000. So, the first two elements of the array will be stored in memory as follows:

1000: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes)
1004: 0 0 0 0 0 0 0 3

Now, %d, *p will print the first 4 bytes starting at 1000, that is 2. When p is incremented, it will go to 1001, as p is a char pointer.

1001: 0 0 0 0 0 0 0 0

So, output will be 0.





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