Can i pass (*ptr) to a function which is taking reference as argument

Question

#include<vector>
using namespace std;

struct x{
    vector<int> y;
};
void magic(struct x& d)
{
    d.y[0] = 5;
}
int main() {
    struct x d;
    d.y = {1,2,3};

    struct x* z = &d;
    magic(*z);

    cout<<z->y[0];
    return 0;
}

Is this code valid and how? Can we pass *z to a function which requires c++ reference.

The function takes a reference to `x`, and you are passing it an `x`. This is similar to `x &d= *z;`. Why do you expect this to not work? — cigien, May 15 '20 at 18:07
...or more generally, why should there be any difference between `d` and `*z` ? — 463035818_is_not_an_ai, May 15 '20 at 18:10
In C, you need to do `struct x d;`. In C++, you only need to do `x d;`. — Eljay, May 15 '20 at 18:11
`#include using namespace std;` - [No, no, no](https://stackoverflow.com/q/31816095/5910058), don't *ever* do that. — Jesper Juhl, May 15 '20 at 18:13

score 0 · Answer 1 · answered May 15 '20 at 18:11

The pointer z points to the object d.

struct x* z = &d;

Dereferencing the pointer you get the lvalue of the object d that is passed as an argument to the function by reference

magic(*z);

So there is no magic.

In fact these two calls

magic( *z );

and

magic( d );

are equivalent. The only difference is that in the first call there is used a more complicated expression.

Can i pass (*ptr) to a function which is taking reference as argument

1 Answers1