C++11构造函数继承和无参数构造函数

Question

这段代码中，为什么A的无参构造函数没有被继承呢？是否有特殊规则阻止继承不带参数的构造函数？

struct A {
    A(void *) {}
    A() {}
};

class B : public A {
public:
    using A::A;
    B(int x) {}
};

void f() {
    B b(1);
    B b2(nullptr);
    B b3; // error
}

clang++ -std=c++11 给出了这个错误，而 g++ -std=c++11 给出了一个基本相似的错误消息:

td.cpp:15:7: error: no matching constructor for initialization of 'B'
    B b3; // error
      ^
td.cpp:9:5: note: candidate constructor not viable: requires single argument 'x', but no arguments
      were provided
    B(int x) {}
    ^
td.cpp:8:14: note: candidate constructor (inherited) not viable: requires 1 argument, but 0 were
      provided
    using A::A;
             ^
td.cpp:2:5: note: inherited from here
    A(void *) {}

Answer 1

相关信息在12.9 [class.inhctor]第3段(加亮):

For each non-template constructor in the candidate set of inherited constructors other than a constructor having no parameters or a copy/move constructor having a single parameter, a constructor is implicitly declared with the same constructor characteristics unless there is a user-declared constructor with the same signature in the complete class where the using-declaration appears. [...]

也就是说，默认构造函数不会被继承，除非它们有一个 [defaulted] 参数。如果它们有默认参数，它们将被继承，但没有默认参数，正如同一段落上的节点所指出的那样:

Note: Default arguments are not inherited. [...]

基本上，这表示默认构造函数不是继承的。