Programming and DS

Question

#include <stdio.h>

int main(){

int a[] = {5,3,7,2,4};

int *p = &a[3];

p -= *p;

printf("%d ",*p);

return 0;

}

 

output is 3.

Why 2 * sizeof(int) is doene.?

Comments

@Abhrajyoti00 @gatecse 

your opinion on this question.

 

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$ p-=*p $ can be written as $p=p-(*p)$.

currently, $p$ is pointing to the third element of the array. $(*p)=2$

So $p=p-2$ is pointing to the second element of the array that is $3$

---

Output is 3 .

1. int a[] = {5, 3, 7, 2, 4};: An array a is declared and initialized with five integers.

2. int *p = &a[3];: A pointer p is declared and assigned the address of the element at index 3 in the array a.

3. p -= *p;: The value at the memory location pointed to by p (which is a[3], i.e., 2) is subtracted from the pointer p. This means p now points to the memory location p - 2.

4. printf("%d ", *p);: The value at the memory location pointed to by p is printed. Since p now points to p - 2, it will print the value at a[1] (assuming zero-based indexing), which is 3.

Answer 1

Because it is an integer array and the command p -= p*(p* = dereferenced value of pointer p = 2) which is equal to p = p – 2, means we are decreasing the address of the pointer by 2*4 byte i.e. 8 bytes.

Answer 2

         It Will Print 3

1. int a[] = {5, 3, 7, 2, 4};: An array a is declared and initialized with values.

2. int *p = &a[3];: A pointer p is declared and initialized with the address of the fourth element of the array (a[3]), which is 2.

3. p -= *p;: This line is equivalent to p = p - *p;. The value at the address p is dereferenced (*p), which is 2. Then, this value is subtracted from the current value of the pointer p. So, p becomes p - 2. After this operation, p points to the second element of the array (a[1]).

4. printf("%d ", *p);: This line prints the value at the current location pointed by p, which is a[1]. So, it prints 3.