#include<stdio.h> void fun(int **p); int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; int *ptr; ptr = &a[0][0]; fun(&ptr); return 0; } void fun(int **p) { printf("%d\n", **p); }
Step 2: int *ptr; The *ptr is a integer pointer variable.
Step 3: ptr = &a[0][0]; Here we are assigning the base address of the array a to the pointer variable *ptr.
Step 4: fun(&ptr); Now, the &ptr contains the base address of array a.
Step 4: Inside the function fun(&ptr); The printf("%d\n", **p); prints the value '1'.
because the *p contains the base address or the first element memory address of the array a (ie. a[0])
**p contains the value of *p memory location (ie. a[0]=1).
Hence the output of the program is '1'
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.