Initializer list don't need parametric constructor?

Question

Hi
I thought classname obj{Val} is initializer list..... So, what is classname obj = {Val}
Does second option i.e. with equal to call implicit constructor ?
#include <iostream>
using namespace std;
class X
{
public:
int x;
explicit X(int x) : x(x){}
};
int main()
{
X a = {10};
X b = a;
cout << a.x << " " << b.x << endl;
return 0;
}
Above code only work if I remove explicit keyword .... Also why it build without no parametric constructor as well ?

Answer 1

this "explicit" makes you unable to copy-initialization. You have to make it "X a(10)" and "X b(a)" so it is direct-initialization. You can use static_cast too, like "X c = (X)5"

Answer 2

About implicit constructor, I'm not sure about that, but implicit constructor (copy constructor in this example) can actually work, try
X a(10); //calling directly constructor
X b = a; // copy constructor
Idk if it works tho

Answer 3

I got your point , but does it not required to have parametric constructor ? Assignment will work as it is generated by default... Does parametric constructor also generated by default due to initializer list ?

Answer 4

Not really, assignment is not generated by default. If you use inline constructor (generated by default), you have to assign values in class declaration like that:
public:
int x {0};
Then you do not need your own parametric constructor.
Is it what were you asking for?