将引用绑定到指针的语法是什么?(所有类型) [英] what's the syntax to bind reference to pointer?(all kinds)

问题描述

这是代码:

int main()
{   
    int* p = nullptr;
    const int* cp = p;
    const int*& ref = cp;
    const int*& ref0 = cp;
    const int*& ref1 = p;//not allowed
    const int*& ref2 = static_cast<const int*&>(p);//not allowed
    int* const & ref3 = p;//allowed. Why?
    return 0;
}

运行它时，我得到了

test.cpp:7:14: error: non-const lvalue reference to type 'const int *' cannot
      bind to a value of unrelated type 'int *'
        const int*& ref1 = p;
                    ^      ~
test.cpp:8:21: error: non-const lvalue reference to type 'const int *' cannot
      bind to a value of unrelated type 'int *'
        const int*& ref2 = static_cast<const int*&>(p); 

我无法理解原因.也许const在这里起作用.当我写const int*&或int* const &时，我不知道const适用于指针或引用.而且const int* const &也使我感到困惑. 我想知道以下项目的语法:

I am having trouble understanding why. Maybe const plays a role here. And when I wrote const int*& or int* const &, I don't know const apply to pointer or reference. And const int* const & confuses me too. I want to know the syntax for the following item:

对const指针的引用.

A reference to a pointer that is const.

对指向const的指针的引用.

A reference to a pointer that points to a const.

对const指针的引用，该指针指向const.

A reference to a pointer that is const and points to a const.

和const参考版本:

And const reference version:

对指针的const引用.

A const reference to a pointer.

对const指针的const引用.

A const reference to a pointer that is const.

指向指向const的指针的const引用.

A const reference to a pointer that points to a const.

const指向const并指向const的指针.

A const reference to a pointer that is const and points to a const.

对不起，我很困惑.

推荐答案

您的p是指向int的指针. const int*&是对const int的指针的引用.由于p是int而不是const int，因此您不能将应该引用指向const int的指针的引用绑定到引用p.

Your p is a pointer to an int. const int*& is a reference to a pointer to a const int. Since p is an int and not a const int, you cannot bind a reference that is supposed to refer to a pointer to a const int to refer to p.

在C ++中读取声明的关键是从内而外"读取它们，即从标识符(名称)开始，然后逐步解决.举个例子:

The key to reading declarations in C++ is to read them "from the inside out", i.e., start from the identifier (name) and work your way out. Take your example:

int* const & ref3;

是对int的const指针的引用.我们从标识符(ref3)开始并向外进行.我们发现的第一件事是&.因此ref3是参考.下一个是const，所以ref3所指的是const.下一个是*，因此引用引用的const是指针.最后，int，我们正在处理对int的const指针的引用.

is a reference to a const pointer to an int. We start at the identifier (ref3) and work outwards. The first thing we find is an &. So ref3 is a reference. Next thing is const, so whatever ref3 refers to is a const something. Next thing is a *, so the const thing the reference refers to is a pointer. Finally int, we're dealing with a reference to a const pointer to an int.

请注意，标识符的两面可能都有东西. 解决问题"时，您必须考虑首先/更强地绑定哪些说明符/运算符，以弄清楚所声明的类型.例如:

Note that you can have stuff happening to both sides of the identifier. When "working your way out", you have to consider which specifiers/operators bind first/more strongly to figure out what type is declared. For example:

int const * a[10];

同样，我们从标识符a开始. []的绑定比*绑定更牢固，因此a是10个元素的数组. a是数组的哪些元素?接下来是*，因此a是10个指针的数组.最后，我们发现a是10个指向const int的指针的数组.请注意，如果结尾处还有一个寂寞的const，则该const会绑定到之前的任何内容.这就是让我们也可以写与int const a等效的const int a的原因.

Again, we start at the identifier a. The [] binds more strongly than *, so a is an array of 10 elements. What are these elements that a is an array of? Next comes a *, so a is an array of 10 pointers to something. In the end, we find that a is an array of 10 pointers to const int. Note that if there's a lonely const left at the end, that const binds to whatever came before it. That's what allows us to also write const int a which is equivalent to int const a.

您还可以使用括号来影响运算符在声明中生效的顺序.例如

You can also use parentheses to influence the order in which operators take effect in a declaration. For example

int const (* a)[10];

将是指向10个const int数组的指针，而不是指向const int的10个指针的数组.

would be a pointer to an array of 10 const int rather than an array of 10 pointers to const int.

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