View source for Pointer qn 1 ← Pointer qn 1

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 is 1000. So, the first two elements of the array will be stored in memory as follows:

1000: 0 2 0 0 0 0 0 0 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at highest address(little-endian))
1004: 0 3 0 0 0 0 0 0

Now, %d, *p. Here, *p is a character which is just a byte. So, *p value here is the byte at address 1000 which is 00 (in hex) and this is passed to printf. In C language, when a char is passed to a function actually it's int value as given by the char code (usually ASCII) is passed. So, here 00 is extended to 0000 and passed to printf. Now, in printf we used %d, and hence the output will be 2.

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.