C Programming - Pointers - Discussion

Discussion Forum : Pointers - Find Output of Program (Q.No. 13)
13.
What will be the output of the program?
#include<stdio.h>

int main()
{
    int arr[3] = {2, 3, 4};
    char *p;
    p = arr;
    p = (char*)((int*)(p));
    printf("%d, ", *p);
    p = (int*)(p+1);
    printf("%d", *p);
    return 0;
}
2, 3
2, 0
2, Garbage value
0, 0
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
87 comments Page 1 of 9.

Badareenath said:   2 decades ago
Can anybody explain this please ?

Ravi said:   2 decades ago
p=arr;//here p is a character pointer. we should not assign int array base address...

DRukus said:   2 decades ago
p is declared as a char pointer but arr is an integer array
the assignment p = arr produces a compile-time error.

Rajadurai said:   2 decades ago
The answer is correct. We can convert int pointer to char pointer and viz versa. *p is apoint to first variable. *p+1 points the next array, so the answer is zero.

Sundar said:   1 decade ago
The char pointer only stores the first byte address of integer. So while increment the pointer the next value is to be printed as 0.

Prateek said:   1 decade ago
p is a char. pointer and it is assigned to complete address of arr array. so when we write (ar+1) it take address next beyond arr array. which having value 0 of defined pointer array.

Sai said:   1 decade ago
Please explain about
p = (int*)(p+1);
printf("%d", *p);

Purushotham said:   1 decade ago
p = arr;
causes compile time error

Viraj said:   1 decade ago
The answer is correct. Assuming that integer requires 2 bytes of storage
the integer array will be stored in memory as

00000010 00000000 00000011 00000000 00000100 00000000

This is because little-endian method is used to store bytes in memory i.e. least significant bytes are stored first.

When a character pointer is incremented by it points to the next byte. Hence the answer.
(1)

Nitin said:   1 decade ago
Well explained Viraj.


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