如何在C ++中正确初始化Struct

Question

Currently running into an issue with this question in my homework. I believe that only 4,7 are incorrect (In Visual Studio they don't throw an error). But i'm honestly not sure why they are the only ones. Since 3 works, I assumed they would work as well but that seems to not be the case. Any advice?

struct A {
    double x;
    A(double x = 1) : x(x) { }
};

struct B {
    double x;
    B(A a = 2.0) : x(a.x) { }
};

struct C {
    double x;
    C(B b = B(3)) : x(b.x) { }
};

int main() {
    A a; // (1)
    A a = 4; // (2)
    B b; // (3)
    B b = 5; // (4)
    B b(a); // (5) (a is an object of class A)
    C c; // (6)
    C c = 6.0; // (7)
    C c(a); // (8) (a is an object of class A)
    C c(b); // (9) (b is an object of class B)
}

Correct ones are:

a) 1-3

b) 1-3,5,9

c) 1-6,8,9

d) 1-7

e) 1-3,5,6,9

f) none

My reasoning:

1) Correct, just default constructor

2)Correct, constructor default or value (4)

3) Correct, default constructor

4) Incorrect, No constructor for an int

5) Correct, exists constructor for object of type A

6) Correct, Default

7)Incorrect, same as 4

8) This one i'm not sure on, there is no constructor for objects of type A, so I would say incorrect

9) Correct, constructor exists.

This is my reasoning in any case, but i'm not sure where i'm going wrong.

Answer 1

The rule is, when converting from a type to another (here from/to int , double , A , B or C ): only one user-provided conversion can be used .

This makes indeed B b = 5; // (4) B b = 5; // (4) invalid since 5 (an int ) needs to be:

• converted to double (first standart-conversion ),
• then to a A (first user-defined conversion ),
• then to a B (second user-defined conversion ).

That last one breaks the rule, and this conversion sequence is not legal.

I'm honestly not sure why they are [ incorrect ].

You can use this rule to check other expressions.

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Finally, you can impress your teacher with std::is_convertible :

std::is_convertible_v<B, C> returns true iff B is convertible to C ( demo ).