显式复制构造函数行为和pratical使用 [英] Explicit copy constructor behavior and pratical uses

问题描述

最近的一个问题让我想知道显式复制构造函数。下面是我在Visual Studio 2005下编译的示例代码：

A recent question got me wondering about explicit copy constructors. Here is a sample code that I tried compiling under Visual Studio 2005 :

struct A
{
    A() {}
    explicit A(const A &) {}
};

// #1 > Compilation error (expected behavior)
A retByValue()
{
    return A();
}

// #2 > Compiles just fine, but why ?
void passByValue(A a)
{
}

int main()
{
    A a;
    A b(a); // #3 > explicit copy construction : OK (expected behavior)
    A c = a; // #4 > implicit copy construction : KO (expected behavior)

    // Added after multiple comments : not an error according to VS 2005.
    passByValue(a);
    return 0;
}

现在提出问题：

• 标准是否允许＃2？如果是，那么描述这种情况的相关部分是什么？


• 你知道显式复制构造函数的任何实际用途吗？

我在 MSDN 具有完全相同的情况，并从主要功能的一个神秘的注释：c被复制（好像很明显）。

I just found a funny link on MSDN with the exact same situation, and a mysterious comment from the main function : "c is copied" (as if it was obvious). As pointed by Oli Charlesworth : gcc does not compile this code and I believe he's right not to.

推荐答案

我相信相关章节的C ++ 03是§12.3.1 2：

I believe the relevant sections of C++03 are §12.3.1 2:

显式构造函数构造对象就像非显式构造函数一样，但只有在直接初始化语法（ 8.5 ）或其中强制转换（ 5.2 .9 ， 5.4 ）。默认构造函数可以是显式构造函数;这样的构造函数将用于执行默认初始化或值初始化（ 8.5 ）。

An explicit constructor constructs objects just like non-explicit constructors, but does so only where the direct-initialization syntax (8.5) or where casts (5.2.9, 5.4) are explicitly used. A default constructor may be an explicit constructor; such a constructor will be used to perform default-initialization or value-initialization (8.5).

和 §8.5 12：

在参数传递，函数返回，抛出异常（15.1 ），处理异常（ 15.3 ）和大括号初始值列表（ 8.5。 1 ）被称为复制初始化，并相当于

The initialization that occurs in argument passing, function return, throwing an exception (15.1), handling an exception (15.3), and brace-enclosed initializer lists (8.5.1) is called copy-initialization and is equivalent to the form

    T x = a;

在新表达式中进行初始化（ 5.3.4 ），static_cast表达式（ 5.2.9 ），功能符号类型转换（ 5.2.3 ），和基本和成员初始值设置（ 12.6.2 ）被称为直接初始化，并且相当于form

The initialization that occurs in new expressions (5.3.4), static_cast expressions (5.2.9), functional notation type conversions (5.2.3), and base and member initializers (12.6.2) is called direct-initialization and is equivalent to the form

    T x(a);

调用 passByValue code>涉及复制初始化，而不是直接初始化，因此应该是一个错误，根据C ++ 03§12.3.1 2。

Calling passByValue(a) involves copy-initialization, not direct-initialization, and thus should be an error, according to C++03 § 12.3.1 2.

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