About two phase lookup and this pointer

Question

In c++ language, I learned about two phase lookup applied on template.
It said compiler checks twice at template.
But I can't understand the correlation between this pointer and two phase lookup.
So please tell me why removing this pointer (commented region) makes error.

#include <iostream>
using namespace std;

template <class T>
class A {
public:
    void f() {
        cout << "f()\n";
    }
};

template <class T>
class B : public A<T> {
public:
    void g() {

        //this->f();
    }
};

int main()
{
    B<int> b;
    b.g();
}

What error do you get? I tried 3 compilers, and they all work. — ChrisMM, Dec 13 '19 at 12:58
do you mean you get an errror when you call `f();` instead of `this->f();` ? It is better to show the code that causes the error and the error. Showing code that needs some modification to get the error always causes confusion — 463035818_is_not_an_ai, Dec 13 '19 at 13:02
Also look at https://stackoverflow.com/questions/7767626/two-phase-lookup-explanation-needed — Peter, Dec 13 '19 at 13:02

score 1 · Answer 1 · answered Dec 13 '19 at 13:00

1

You need to look at the type of this. It's a B<T>* in your example, which means it depends on T. Therefore the name lookup of f in this->f is done in the second phase, where it's known that T==int and thus this is a B<int>*.

answered Dec 13 '19 at 13:00

MSalters

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score 0 · Answer 2 · answered Dec 13 '19 at 13:02

With

f();

f does not depend on a template parameter and thus looked up only in the template definition context. With

this->f();

f does dependent on a template parameter (because this does as the parent class does) and thus looked up in the template instanciation context.

About two phase lookup and this pointer

2 Answers2