据我理解,例如阅读此处的内容,派生类的构造函数不会调用其虚拟基类的构造函数。
这里是我制作的一个简单示例:
class A {
protected:
A(int foo) {}
};
class B: public virtual A {
protected:
B() {}
};
class C: public virtual A {
protected:
C() {}
};
class D: public B, public C {
public:
D(int foo, int bar) :A(foo) {}
};
int main()
{
return 0;
}
出于某些原因,构造函数B::B()
和C::C()
试图初始化 A
(在我的理解中,此时应该已经被D
初始化):
$ g++ --version
g++ (GCC) 10.2.0
Copyright (C) 2020 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
$ g++ test.cpp
test.cpp: In constructor ‘B::B()’:
test.cpp:8:13: error: no matching function for call to ‘A::A()’
8 | B() {}
| ^
test.cpp:3:9: note: candidate: ‘A::A(int)’
3 | A(int foo) {}
| ^
test.cpp:3:9: note: candidate expects 1 argument, 0 provided
test.cpp:1:7: note: candidate: ‘constexpr A::A(const A&)’
1 | class A {
| ^
test.cpp:1:7: note: candidate expects 1 argument, 0 provided
test.cpp:1:7: note: candidate: ‘constexpr A::A(A&&)’
test.cpp:1:7: note: candidate expects 1 argument, 0 provided
test.cpp: In constructor ‘C::C()’:
test.cpp:13:13: error: no matching function for call to ‘A::A()’
13 | C() {}
| ^
test.cpp:3:9: note: candidate: ‘A::A(int)’
3 | A(int foo) {}
| ^
test.cpp:3:9: note: candidate expects 1 argument, 0 provided
test.cpp:1:7: note: candidate: ‘constexpr A::A(const A&)’
1 | class A {
| ^
test.cpp:1:7: note: candidate expects 1 argument, 0 provided
test.cpp:1:7: note: candidate: ‘constexpr A::A(A&&)’
test.cpp:1:7: note: candidate expects 1 argument, 0 provided
我确定有些非常基础的东西我可能误解了或者做错了,但我无法弄清楚。