在 C & 中指向带有 const 限定符的数组的指针C++

2022-01-23 00:00:00 arrays pointers c constants c++

考虑以下程序:

int main()
{
    int array[9];
    const int (*p2)[9] = &array;
}

它在 C++ 中编译良好(参见现场演示 这里)但在 C 中编译失败.默认情况下，GCC 提供以下警告.(查看现场演示这里).

It compiles fine in C++ (See live demo here) but fails in compilation in C. By default GCC gives following warnings. (See live demo here).

prog.c: In function 'main':
prog.c:4:26: warning: initialization from incompatible pointer type [enabled by default]
     const int (*p2)[9] = &array;

但如果我使用 -pedantic-errors 选项:

But If I use -pedantic-errors option:

gcc -Os -s -Wall -std=c11 -pedantic-errors -o constptr constptr.c

它给了我以下编译器错误

it gives me following compiler error

constptr.c:4:26: error: pointers to arrays with different qualifiers are incompatible in ISO C [-Wpedantic]

为什么它在 C 中编译失败，而在 C++ 中却没有?什么C&C++ 标准对此有何规定?

Why it fails in compilation in C but not in C++? What C & C++ standard says about this?

如果我在数组声明语句中使用 const 限定符，它在 C 中也可以正常编译.那么，在上面的程序中发生了什么?

If I use const qualifier in array declaration statement it compiles fine in C also. So, what is happening here in above program?

推荐答案

GCC-gnu

在 GNU C 中，指向带有限定符的数组的指针与指向其他限定类型的指针类似.例如，int (*)[5] 类型的值可用于初始化 const int (*)[5] 类型的变量.这些类型在 ISO C 中是不兼容的，因为 const 限定符正式附加到数组的元素类型而不是数组本身.

In GNU C, pointers to arrays with qualifiers work similar to pointers to other qualified types. For example, a value of type int (*)[5] can be used to initialize a variable of type const int (*)[5]. These types are incompatible in ISO C because the const qualifier is formally attached to the element type of the array and not the array itself.

C 标准说(部分:§6.7.3/9):

C standard says that (section: §6.7.3/9):

如果数组类型的规范包括任何类型限定符，则元素类型是这样限定的，不是数组类型.[...]

If the specification of an array type includes any type qualifiers, the element type is so- qualified, not the array type.[...]

现在看看 C++ 标准(第 3.9.3/5 节):

Now look at the C++ standard (section § 3.9.3/5):

[...] 应用于数组类型的 Cv 限定符附加到底层元素类型，因此符号cv T"，其中 T 是一个数组type，指的是一个数组，其元素是如此限定的.元素经过 cv 限定的数组类型也被认为具有与其元素相同的 cv 限定.[ 例子:

[...] Cv-qualifiers applied to an array type attach to the underlying element type, so the notation "cv T," where T is an array type, refers to an array whose elements are so-qualified. An array type whose elements are cv-qualified is also considered to have the same cv-qualifications as its elements. [ Example: 

 typedef char CA[5];
 typedef const char CC;
 CC arr1[5] = { 0 };
 const CA arr2 = { 0 };

arr1和arr2的类型都是array of 5 const char"，并且数组类型被认为是是 const 限定的.――结束示例]

The type of both arr1 and arr2 is "array of 5 const char," and the array type is considered to be const- qualified. ―endexample]

因此，初始化

const int (*p2)[9] = &array;

将类型指向int的数组[9]的指针分配给指向constint的数组[9]的指针.这与将 int * 分配给 const int * 不同，其中 const 直接应用于指针指向的 object 类型.这不是 const int(*)[9] 的情况，在 C 中，const 应用于数组对象的元素而不是指针指向的对象到.这使得上述初始化不兼容.

is assignment of type pointer to array[9] of int to pointer to array[9] of const int. This is not similar to assigning int * to a const int * where const is applied directly to the object type the pointer points to. This is not the case with const int(*)[9] where, in C, const is applied to the elements of the array object instead of the object the pointer points to. This makes the above initialization incompatible.

此规则在 C++ 中已更改.由于 const 应用于数组对象本身，因此赋值是在相同类型 pointer to const array[9] of int 而不是类型 指向 int 的数组 [9] 的指针和 指向 const int 的数组 [9] 的指针.

This rule is changed in C++. As const is applied to array object itself, the assignment is between same types pointer to const array[9] of int instead of type pointer to array[9] of int and pointer to array[9] of const int.

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