C++ 如何在构造函数中初始化 const 字段？
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原文地址: http://stackoverflow.com/questions/1423696/
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How to initialize a const field in constructor?
提问by puccio
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?
想象一下，我有一个 C++ 类 Foo 和一个类 Bar，它们必须使用传递 Foo 指针的构造函数创建，并且该指针旨在在 Bar 实例生命周期中保持不变。正确的做法是什么？
In fact, I thought I could write like the code below but it does not compile..
事实上，我以为我可以像下面的代码那样写，但它不能编译..
class Foo;
class Bar {
public:
Foo * const foo;
Bar(Foo* foo) {
this->foo = foo;
}
};
class Foo {
public:
int a;
};
Any suggestion is welcome.
欢迎任何建议。
回答by Jim Buck
You need to do it in an initializer list:
您需要在初始化列表中执行此操作：
Bar(Foo* _foo) : foo(_foo) {
}
(Note that I renamed the incoming variable to avoid confusion.)
（请注意，我重命名了传入变量以避免混淆。）
回答by SingleShot
I believe you must do it in an initializer. For example:
我相信您必须在初始化程序中执行此操作。例如：
Bar(Foo* foo) : foo(foo) {
}
As a side note, if you will never change what foo points at, pass it in as a reference:
作为旁注，如果您永远不会更改 foo 指向的内容，请将其作为参考传递：
Foo& foo;
Bar(Foo& foo) : foo(foo) {
}
回答by ezpz
Initializing const members and other special cases (such a parent classes) can be accomplished in the initializer list
初始化 const 成员和其他特殊情况（如父类）可以在初始化列表中完成
class Foo {
private:
const int data;
public:
Foo(int x) : data(x) {}
};
Or, similarly, for parent initialization
或者，类似地，对于父初始化
class Foo {
private:
int data;
public:
Foo(int x) : data(x) {}
};
class Bar : Foo {
public:
Bar(int x) : Foo(x) {}
};
回答by KeithB
You need to initialize foo in the initializer list.
您需要在初始化列表中初始化 foo。
class Bar {
Foo* const foo;
public:
Bar(Foo* f) : foo(f) {...}
};
回答by AraK
Use a reference:
使用参考：
Foo& foo;
Bar(Foo& f) : foo(f) { }
You can then refer to fooeasily in Bar:
然后您可以foo轻松地 参考Bar：
foo.doSomething();
回答by John Ledbetter
try: Bar(Foo* xfoo) : foo(xfoo) {}
尝试： Bar(Foo* xfoo) : foo(xfoo) {}
