why does *A is required i mean why is pointer needed here....plz answer....

Question

#include<stdio.h>

void printArray(int *A, int n){ //i am talking about this *a
for (int i = 0; i < n; i++)
{
printf("%d ", A[i]);
}
printf("\n");
}

int main(){

int A[] = {1, 2, 5, 6, 12, 54, 625, 7, 23, 9, 987};
int n = 11;

printArray(A, n);

return 0;
}

Answer 1

When you assign an array to a variable, you actually assign the pointer to the first element of the array. To test this, run the following lines
int arr[] = { 1, 2, 3 };
cout << *arr;
// Outputs 1 because arr is actually pointer to first element (1) of the array

So when you pass an array to a function, and in your case, when you pass `A` to `printArray`, actually the pointer to the first element of the array is passed. That is why you are able to recieve an array in a function where `int *` is required.

NOTE: it is not compulsory to specify the type of parameter `A` as `int *`. You can also use `int A[]`. So your function can also be defined as
void printArray(int A[], int n) {
......
}