什么是'&'在C ++声明中？ [英] What does '&' do in a C++ declaration?

问题描述

我是一个C的家伙，我试图理解一些C ++代码。我有以下函数声明：

I am a C guy and I'm trying to understand some C++ code. I have the following function declaration:

int foo(const string &myname) {
  cout << "called foo for: " << myname << endl;
  return 0;
}

函数签名如何与等价的C：

How does the function signature differ from the equivalent C:

int foo(const char *myname)

b $ b

使用 string * myname vs string& myname 有区别吗？ C中的& 和C中的 * 之间有什么区别，以指示指针？

Is there a difference between using string *myname vs string &myname? What is the difference between & in C++ and * in C to indicate pointers?

类似地：

const string &GetMethodName() { ... }

& 在这里做什么？是否有一些网站解释了& 在C和C ++中的用法不同？

What is the & doing here? Is there some website that explains how & is used differently in C vs C++?

推荐答案

&表示引用而不是指向对象的指针（在您的情况下是常量引用）。

The "&" denotes a reference instead of a pointer to an object (In your case a constant reference).

具有

foo(string const& myname)

过

foo(string const* myname)

你保证myname是非空的，因为C ++不允许NULL引用。

is that in the former case you are guaranteed that myname is non-null, since C++ does not allow NULL references. Since you are passing by reference, the object is not copied, just like if you were passing a pointer.

你的第二个例子：

const string &GetMethodName() { ... }

b $ b

允许你返回一个常量引用，例如，一个成员变量。如果您不希望返回副本，并且再次确保返回的值非空，这将非常有用。例如，以下内容允许您直接，只读访问：

Would allow you to return a constant reference to, for example, a member variable. This is useful if you do not wish a copy to be returned, and again be guaranteed that the value returned is non-null. As an example, the following allows you direct, read-only access:

class A
{
  public:
  int bar() const {return someValue;}
  //Big, expensive to copy class
}

class B
{
public:
 A const& getA() { return mA;}
private:
 A mA;
}
void someFunction()
{
 B b = B();
 //Access A, ability to call const functions on A
 //No need to check for null, since reference is guaranteed to be valid.
 int value = b.getA().bar(); 
}

您必须注意不返回无效引用。
编译器将很乐意编译以下内容（取决于您的警告级别和处理警告的方式）

You have to of course be careful to not return invalid references. Compilers will happily compile the following (depending on your warning level and how you treat warnings)

int const& foo() 
{
 int a;

 //This is very bad, returning reference to something on the stack. This will
 //crash at runtime.
 return a; 
}

基本上，你有责任确保你返回的引用实际上是有效的。

Basically, it is your responsibility to ensure that whatever you are returning a reference to is actually valid.

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