代码比lambda简单，在构造函数中使用一个输出参数函数初始化一个const成员 [英] Code simpler than lambda for a call in constructor that uses an output parameter function for initializing a const member

问题描述

在标题中，我有

class CSomeClass
{
    const GUID m_guid;

public:
    CSomeClass();
///...
}

/ p>

And in the source file

CSomeClass::CSomeClass()
    , m_guid(
        []() {
        GUID g;
        ::CoCreateGuid(&g);
        return g;
        }()
    )
{
}

正如你所知道的Guids可以被用作标识，而不是要改变。给定 :: CocreateGuid（）函数提供了我想要的作为输出参数，而不是返回它，我不能直接使用一个简单的调用函数来初始化m_guid成员字段，即常数。

As you know Guids can be used as identifications not meant to be changed. Given the ::CocreateGuid() function provides what I want as an output parameter, instead of returning it, I cannot use directly a simple call to the function for initializing the m_guid member field, that is constant.

所以，它的常量的一个结果是，它必须在初始化列表中的开始括号之前初始化，因此不能简单地分配一个 :: CocreateGuid（）在构造函数体中。

So, a consequence of its constness, is that it must be initialized before the opening bracket in initializer list, and therefore not be simply assigned with a call to ::CocreateGuid() in the constructor body.

有比这个lambda表达式更简单的方法来初始化它吗？ p>

Is there a simpler way to initialize it than this lambda expression?

推荐答案

当lambda表达式正确时，我将使用一个辅助函数：

When the lambda expression is correct, I would use a helper function for that:

GUID create_guid()
{
    GUID g;
    ::CoCreateGuid(&g);
    return g;
}

CSomeClass::CSomeClass() : m_guid(create_guid()) {}

此外， create_guid（）本身有一个意义，可以重用（即使使一个实现细节是可能的/正确的）。

In addition, create_guid() has a meaning by itself and could be reused (even if making it a implementation detail is possible/correct).

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