我的理解,比如阅读this, https://www.cprogramming.com/tutorial/virtual_inheritance.html是派生类的构造函数不调用其虚拟基类的构造函数。
这是我做的一个简单的例子:
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;
}
由于某种原因,构造函数B::B()
and C::C()
正在尝试初始化A
(按照我的理解,它应该已经被初始化了D
在此刻):
$ 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
我确信我误解了或者做错了一些非常基本的事情,但我不知道是什么。