It was my whole life question.

Question

#include <iostream>

using namespace std;

void foo(int *ptr, int *secPtr){
int *sum = new int;
*sum = *ptr + *secPtr;

cout << *sum;
}

int main(){
int *a = new int;
*a = 5;
int *b = new int;
*b = 6;

foo(*a, *b);

return 0;
}

Answer 1

foo(&a, &b);
doesn't work as it creates arguments of type:
int **ptr
pointer to pointer of integer so you would get a type mismatch.

Answer 2

Your foo function is looking for two pointers to integers (int *ptr, int *secPtr). Both a and b are declared as pointers to integers with:
int *a = new int;
int *b = new int;
Therefore, they are the correct type to be passed into foo so this works:
foo(a, b);
You could also declare these:
int c,d;
and pass them as:
foo(&c, &d);
making pointers by using the address operator.

Answer 3

See this

https://code.sololearn.com/cGH8SIg7qp0H/?ref=app

Answer 4

I think i understood: In this line: foo(*a, *b), the value of a and b are exactly variables(5, 6) and you can't pass them to pointers as pointers only accept an adress of variables, thus you can't pass them by reference all the time.

Answer 5

The problem is in int main function, foo can't be called, why?
compiler says: no matching function to call for foo();

Answer 6

Meet Mehta, nope

Answer 7

Meet Mehta, this works: (a, b), but i can't quite understand why🤔