View source for Pointer qn 1 ← Pointer qn 1

==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

==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
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 lowest 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 02 (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 02 is sign-extended to 0002 and passed to printf. Now, in printf we used %d, and hence the output will be 2. 

 When p is incremented, it will become to 1001, as p is a char pointer (for pointer addition, sizeof datatype pointed to is added and sizeof char is 1). So, now the value of *p is 00 and as explained above printf will print 0 for this. 
 1001: 0 0 0 0 0 0 0 0

So, output will be 0. 

On a big-endian machine output will be 0 0. So, correct answer should be architecture dependent. 

{{Template:FBD}}



[[Category:Coding Questions]]