Yes,this behavIoUr is intended,according to §13.3.1.7 Initialization
by list-initialization

When objects of non-aggregate class type T are list-initialized (8.5.4),overload resolution selects the constructor in two phases:

— Initially,the candidate functions are the initializer-list constructors (8.5.4) of the class T and the argument list consists of
the initializer list as a single argument.

— If no viable initializer-list constructor is found,overload resolution is performed again,where the candidate functions are all
the constructors of the class T and the argument list consists of the
elements of the initializer list.

在gcc我试过你的例子.我得到这个错误：

error: call of overloaded 'A(<brace-enclosed initializer list>)' is ambiguous

如果我使用三套大括号,gcc会停止抱怨.即：

#include <iostream>
#include <vector>
#include <initializer_list>

struct A {
    A (std::initializer_list<int> il) { 
        std::cout << "First." << std::endl;
    }
    A (std::initializer_list<std::initializer_list<int>> il) { 
        std::cout << "Second." << std::endl;
    }
};

int main()
{
    A a{0}; // first
    A a{{0}}; // compile error
    A a2{{{0}}}; // second
    A a3{{{{0}}}}; // second
}

试图镜像向量的构造函数,这里是我的结果：

#include <iostream>
#include <vector>
#include <initializer_list>

struct A {
    A (std::initializer_list<int> il) { 
        std::cout << "First." << std::endl;
    }
    explicit A (std::size_t n) {
        std::cout << "Second." << std::endl;
    }
    A (std::size_t n,const int& val) {
        std::cout << "Third." << std::endl;
    }
    A (const A& x) {
        std::cout << "Fourth." << std::endl;
    }
};

int main()
{
    A a{0};
    A a2{{0}};
    A a3{1,2,3,4};
    A a4{{1,4}};
    A a5({1,4});
    A a6(0);
    A a7(0,1);
    A a8{0,1};
}

main.cpp:23:10: warning: braces around scalar initializer

    A a2{{0}};

         ^~~

1 warning generated.

First.
First.
First.
First.
First.
Second.
Third.
First.