当您声明“指针类型"时，const是什么?形式(函数)参数“像const数组"? [英] What is const when you declare a "pointer type" formal (function) parameter "like an const array"?

问题描述

当您声明指针类型"形式(函数)参数像const数组"时，const是什么?

What is const when you declare a "pointer type" formal (function) parameter "like an const array"?

即:如果T是类型标识符，则为:

I.e.: If T is a type identifier, then is:

void f(const T arr[]);

等同于:

void f(const T * const arr);

或

void f(const T * arr);

或

void f(T * const arr);

又如何解密C/C ++标准来找到这种等效物? :-)

And how can one to decrypt the C/C++ standards to find this equivalents? :-)

关于上述声明的含义，C和C ++之间以及标准的不同形式之间是否存在差异? (对于C-符合ANSI C89和更高版本的标准)

Are there differences between C and C++ and between the different incarnations of the standards regarding the meaning of the declarations above? (for C - the standards of ANSI C89 and later are relevant)

推荐答案

对于C ++(目前问题也被标记为C):

For C++ (at the moment the question is also tagged as C):

这里涉及三种不同的等价形式. [1] const T arr[]形式参数类型衰减为T const* arr，因此在形式参数类型的上下文中是等效的. [2] T const* arr等效于const T* arr，这是表示它的另一种语法. [3]在函数类型的上下文中，T const* arr中的形式参数类型等效于T const* const arr，因为出于确定函数类型的目的而舍弃了顶级const. IE.以这种方式不同的两个函数声明表示相同的函数.

There are three different kinds of equivalence involved here. [1] The const T arr[] formal argument type decays to T const* arr, which therefore is equivalent in the context of formal argument type. [2] T const* arr is equivalent to const T* arr, which is another syntax expressing the same. [3] In the context of function type the formal argument type in T const* arr is equivalent to T const* const arr, because top level const is discarded for the purpose of determining function type. I.e. two function declarations that differ in this way denote the same function.

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