使用“&” C ++中函数名称之前的运算符 [英] Use of '&' operator before a function name in C++

问题描述

参考定义对象的备用名称。引用类型引用
是另一种类型。我们通过编写形式为& d ，
的声明符来定义引用类型，其中 d 是

A reference defines an alternative name for an object. A reference type "refers to" another type. We define a reference type by writing a declarator of the form &d, where d is the name being declared.

接下来的事情是引用不是对象。相反，引用只是已经存在的对象的另一个名称。因此，我们将使用这些引用逐个传递参数，以便直接影响实际参数。

The next thing is a reference is not an object. Instead, a reference is just another name for an already existing object. So we'll use these references to pass the parameter by reference so that it directly effect the actual parameters.

问题：
当我们在功能名称之前使用引用（& ）时会发生什么？

Question: What happens when we use a reference (&) before a function name?

我有点困惑，因为我认为它将返回return的别名（变量名）。还是我错了？。

I'm a little bit confused, as in my opinion it will return the alias of return (variable name). Or am I wrong?.

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推荐答案

在 C ++中，当在与它关联的函数的声明中的函数名称之前使用引用符号（& ）

In C++, when the ref-sign (&) is used before the function name in the declaration of a function it is associated with the return value of the function and means that the function will return by reference.

int& foo(); // Function will return an int by reference.

在声明上下文中未使用时，将ref符号放在函数名称之前会导致调用 address-of 运算符返回函数的地址。这可以用于例如创建指向函数的指针。

When not used within a declaration context, putting the ref-sign before a function name results in calling the address-of operator returning the address of the function. This can be used to e.g. create a pointer to a function.

// Some function.
int sum(int a, int b) {
    return a + b;
}

int main() {
    // Declare type func_ptr_t as pointer to function of type int(int, int).
    using func_ptr_t = std::add_pointer<int(int, int)>::type;

    func_ptr_t func = &sum; // Declare func as pointer to sum using address-of.
    int sum = func(1, 2);   // Initialize variable sum with return value of func.
}

在 C 中，唯一使用& 用于地址运算符。用C语言存在不引用。

In C, the only use of & is for the address-of operator. References does not exist in the C language.

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