从没有默认构造函数的类继承的类
现在我有一个从 B 类继承的 A 类，并且 B 没有默认构造函数.我正在尝试为 A 创建一个构造函数，它与 B 的构造函数
Right now I have a class A that inherits from class B, and B does not have a default constructor. I am trying the create a constructor for A that has the exact same parameters for B's constructor
struct B {
int n;
B(int i) : n(i) {}
};
struct A : B {
A(int i) {
// ...
}
};
但我明白了:
error: no matching function for call to ‘B::B()’
note: candidates are: B::B(int)
我该如何解决这个错误?
How would I fix this error?
推荐答案
构造函数应该是这样的:
The constructor should look like this:
A(int i) : B(i) {}
冒号后面的位表示使用其int构造函数初始化该对象的B基类子对象，值为i".
The bit after the colon means, "initialize the B base class sub object of this object using its int constructor, with the value i".
我猜您没有为 B 提供初始化程序，因此默认情况下编译器会尝试使用不存在的无参数构造函数对其进行初始化.
I guess that you didn't provide an initializer for B, and hence by default the compiler attempts to initialize it with the non-existent no-args constructor.
相关文章