CODEWITHSUNDEEP
c++initializationinitializer-list
madu
madu

Reputation: 5450

Strange constructor behavior

I am using Visual Studio 2013, and this is what I'm trying to figure out:

#include <iostream>
#include <vector>
using namespace std;

class A
{
public:
    int x = 1;
    bool y = true;

    A(int _x, bool _y) : x(_x), y(_y)
    {
        cout << "Constructor #1" << endl;
    }

    A(std::initializer_list<int> init_list)
    {
        cout << "Constructor #2" << endl;
    }
};


int main(int argc, char** argv)
{
    A Aobj1(10, false);
    A Aobj2(20, "false");
    A Aobj3{30, false};
    A Aobj4{40, "false"};

    return 0;
}

The output is:

Constructor #1
Constructor #1
Constructor #2
Constructor #1

What is some background on the logic of the compiler?

Upvotes: 4

Views: 158

Answers (2)

Jesper Juhl
Jesper Juhl

Reputation: 31447

"false" is a const char[6] which decays to const char* and pointers are implicitly convertible to bool. That's why the compiler can call your constructor taking a int and a bool, since the compiler is allowed to perform one implicit conversion to make a function argument match.

Upvotes: 7

songyuanyao
songyuanyao

Reputation: 172864

Now the second construction of Aobj2(int, string) calling constructor #1 is strange. How is the compiler calling the (int, bool) constructor with a string argument?

More precisely, "false" is of type const char[6] and could decay to const char*, i.e. a pointer, and then could implicitly convert to bool.

A prvalue of integral, floating-point, unscoped enumeration, pointer, and pointer-to-member types can be converted to a prvalue of type bool.

The value zero (for integral, floating-point, and unscoped enumeration) and the null pointer and the null pointer-to-member values become false. All other values become true.

For the 4th case, the parameter std::initializer_list<int> can't match the argument 40, "false", because there're no implicit conversion converting pointer to int. Then the constructor taking int, bool is selected, because "false" could be converted to bool implicitly as the above explained.

Upvotes: 8

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