Geeksforgeeks

Question

// C program to illustrate sizes of // pointer of array #include<stdio.h> int main() {     int arr[] = { 3, 5, 6, 7, 9 };     int *p = arr;           printf("p = %p\n", p);     printf("*p = %d\n", *p);           printf("sizeof(p) = %lu\n", sizeof(p));     return 0; }

Comments

The output shows sizeof(p)=8.But how? Since p is a pointer to array shouldn't it be 4 bytes?

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I didn't get why printf("p = %p\n", p); <--this instruction not produced any output??

I checked output is as:

*p=3                                  // since, printf("*p = %d\n", *p);

sizeof(p) = 8                      //since, printf("sizeof(p) = %lu\n", sizeof(p));

Answer 1

p = address of pointer p. (%p is used to print the address of pointer)

*p = 3 (it is the value at address which is stored in p) or (the data pointed by pointer p)

sizeof(p) = 8 (because it is the size of address stored in p. it is not the size of array)

• address stored in p is of 8 bytes in 64 bit

• address stored in p is of 4 bytes in 32 bit