C ++ 98/03参考崩溃和cv限定符 [英] C++98/03 reference collapsing and cv qualifiers

问题描述

下面的代码编译（gcc 4.7.2或icc 13）并生成1 2输出。这意味着 const 限定符被丢弃，即， f 具有参数类型 int& 。

The code below compiles (gcc 4.7.2 or icc 13) and produces "1 2" output. Which means that const qualifier is dropped, i. e., f<int&> has the parameter type int&.

为什么会发生？据我理解，根据§14.3.1.4：

Why does it happen? As I understand, according to §14.3.1.4:

如果模板参数 T c c c c c T 创建 cv12 S cv12 是cv属性 cv1 和 cv2 的联合。忽略冗余cv-quali文件。

If a template-argument for a template-parameter T names a type "reference to cv1 S", an attempt to create the type "reference to cv2 T" creates the type "reference to cv12 S", where cv12 is the union of the cv-qualifiers cv1 and cv2. Redundant cv-qualifiers are ignored.

const 不应删除。这里是代码：

const should not be dropped. Here is the code:

#include <iostream>
using namespace std;

template <typename T>
void f(const T& t)
{
    t++;
}

int main()
{
    int a = 1;

    cout << a;
    f<int&>(a);
    cout << ' ' << a << endl;

    return 0;
}

推荐答案

似乎起源于：

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2005/n1770.html

14.3.1 - 模板类型参数

14.3.1 - Template type arguments

-4-如果模板参数T名称的模板参数类型lvalue-reference to cv1 S，尝试创建类型（lvalue或rvalue）对cv2 T的引用创建类型lvalue-reference to cv12 S，其中cv12是cv-qualifiers cv1和cv2。如果模板参数名称类型为rvalue-reference to cv1 S，则尝试创建类型lvalue-reference to cv2 T将创建类型lvalue-reference to cv12 S如果模板参数名称类型为rvalue-reference to cv1 S，则尝试创建类型rvalue-reference to cv2 T将创建类型rvalue-reference to cv12 S

-4- If a template-argument for a template-parameter T names a type "lvalue-reference to cv1 S," an attempt to create the type "(lvalue or rvalue) reference to cv2 T" creates the type "lvalue-reference to cv12 S," where cv12 is the union of the cv-qualifiers cv1 and cv2. If the template-argument names a type "rvalue-reference to cv1 S," an attempt to create the type "lvalue-reference to cv2 T" creates the type "lvalue-reference to cv12 S." If the template-argument names a type "rvalue-reference to cv1 S," an attempt to create the type "rvalue-reference to cv2 T" creates the type "rvalue-reference to cv12 S." Redundant cv-qualifiers are ignored.

这里是2118，其中引用了引号：

Here is 2118 where the quote has been struck out:

http：// www .open-std.org / jtc1 / sc22 / wg21 / docs / papers / 2006 / n2118.html

14.3。 1 - 模板类型参数

14.3.1 - Template type arguments

-4-如果模板参数T的模板参数命名类型cv1 S引用即对类型A的引用，尝试创建类型对cv2 T的引用对于cv T的引用创建类型对cv12 S的引用，其中cv12是cv限定符cv1和cv2的联合。忽略冗余cv限定符lvalue-reference to A，而尝试创建类型rvalue-reference to cv T则创建类型T.

-4- If a template-argument for a template-parameter T names a type "reference to cv1 S" that is a reference to a type A, an attempt to create the type "reference to cv2 T" "lvalue-reference to cv T" creates the type "reference to cv12 S", where cv12 is the union of the cv-qualifiers cv1 and cv2. Redundant cv-qualifiers are ignored "lvalue-reference to A", while an attempt to create the type "rvalue-reference to cv T" creates the type T.

您所引用的内容似乎是过时的措辞。

What you are quoting seems to be obsolete wording.

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