指针在C和const修饰词阵列; C ++ [英] Pointer to array with const qualifier in C & C++

问题描述

考虑下面的程序：

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

它编译用C ++罚款（这里见现场演示），但未能在汇编C.默认情况下GCC给出以下警告。 （这里见现场演示）。

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;

不过，如果我使用 -pedantic-错误选项：

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]

为什么在编译失败在C，但不是在C ++？什么C＆放大器; C ++标准说这个呢？

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

如果我使用常量数组声明语句预选赛在C编译罚款也。那么，什么是发生在这里上面的程序？

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，指针与数组工作预选赛类似于指向其他合格的类型。例如，类型的值 INT（*）[5] 可用于初始化类型 const int的（*）的变量[5] 。 这些类型在ISO C不兼容，因为常量预选赛正式连接到阵列的元素类型，而不是该数组本身。

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):

如果一个数组类型的规范包括任何类型的预选赛中，元素类型为所谓合格的不是数组类型即可。[...]

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

现在看一下C ++标准（部分§3.9.3 / 5）：

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

[...]应用到阵列式CV-预选赛附加到底层的元素类型，所以符号 CV牛逼，其中 T 是一个数组类型，指的是一个数组，其元素是如此的限定。 数组类型的元素是CV-资格也被认为具有相同的CV-资格的元素。 [示例的：

[...] 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 };

  
  

该型两种 ARR1 和 ARR2 是5为const char，数组和数组类型被认为是合格的const- 。 -endexample]

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;  

的类型是分配的指针数组[9]的 INT 的到的指针数组[9]为const INT 的。这不是类似于分配为int * 到 const int的* ，其中常量直接施加到对象类型的指针指向。这是不是与[9] 其中，在C，常量应用于 const int的（*）的情况下该阵列的元件对象，而不是该对象的指针指向。这使得上述的初始化不相容。

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.

这个规则在C ++中的变化。由于常量被应用到数组对象本身，任务是常量数组同类型的指针之间[9]的 INT 的替代型的指针数组[9]的 INT 的和的指针数组[9]为const INT 的

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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