返回一个临时的引用 [英] returning a reference to temporary

问题描述

我知道返回一个临时的引用是非法的，但这里是我的问题：

I understand that it's illegal to return a reference to a temporary, but here's my problem:

const stringSet & Target::dirList( const dirType type ) const
{
    switch( type )
    {
        case SOURCE_DIR:
            return m_sourceDirs;
        case HEADER_DIR:
            return m_headerDirs;
        case RESOURCE_DIR:
            return m_resourceDirs;
        default:
            return stringSet(); // PROBLEM HERE!
    }
}

前三个选项返回一个const引用 stringSet 数据成员。我应该对默认情况做什么？如果我离开了，编译器（GCC的 -Wall -Wextra -pedantic ）抱怨，我不想要，因为这些选项倾向于抓住我的床设计选择）

The three first three options return a const reference to a stringSet data member. What should I do for the default case? If I leave it out, the compiler (GCC with -Wall -Wextra -pedantic) complains and I don't want it to because those options tend to catch my bed design choices in the most odd of ways :)

谢谢！

推荐答案

保持默认设置为成员...并返回对它的引用。这当然是如果默认情况下理论上可能。如果不是，抛出异常，不返回任何东西。

Keep a default set as a member too... and return a reference to it. That is of course if the default case is theoretically possible. If it is not, throw an exception and don't return anything.

default:
   throw invalid_argument_exception();

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