如何在构造函数中初始化 const 字段?
想象一下，我有一个 C++ 类 Foo 和一个类 Bar，它们必须用一个构造函数来创建，在该构造函数中传递了一个 Foo 指针，并且这个指针在 Bar 实例生命周期中保持不变.正确的做法是什么?
Imagine I have a C++ class Foo and a class Bar which has to be created with a constructor in which a Foo pointer is passed, and this pointer is meant to remain immutable in the Bar instance lifecycle. What is the correct way of doing it?
事实上，我以为我可以像下面的代码那样写，但它不能编译..
In fact, I thought I could write like the code below but it does not compile..
class Foo;
class Bar {
public:
Foo * const foo;
Bar(Foo* foo) {
this->foo = foo;
}
};
class Foo {
public:
int a;
};
欢迎提出任何建议.
推荐答案
你需要在初始化列表中进行:
You need to do it in an initializer list:
Bar(Foo* _foo) : foo(_foo) {
}
(请注意，我重命名了传入变量以避免混淆.)
(Note that I renamed the incoming variable to avoid confusion.)
相关文章