警告：返回引用临时 [英] warning: returning reference to temporary

问题描述

我有这样的功能

const string &SomeClass::Foo(int Value)
{
    if (Value < 0 or Value > 10)
        return "";
    else
        return SomeClass::StaticMember[i];
}

我得到警告： code>。这是为什么？我认为函数返回的两个值（引用const char *和引用静态成员）不能是临时的。

I get warning: returning reference to temporary. Why is that? I thought the both values the function returns (reference to const char* "" and reference to a static member) cannot be temporary.

推荐答案

p>这是一个不必要的隐式转换发生的例子。 不是一个std :: string，所以编译器试图找到一种方法将其变成一个。通过使用字符串（const char * str）构造函数，它在这个尝试中成功。
现在已经创建了一个std :: string的临时实例，它将在方法调用结束时被删除。因此，引用一个在方法调用后不再存在的实例显然不是一个好主意。
我建议您将返回类型更改为const字符串或将存储在SomeClass的成员或静态变量中。

This is an example when an unwanted implicit conversion takes place. "" is not a std::string, so the compiler tries to find a way to turn it into one. And by using the string( const char* str ) constructor it succeeds in that attempt. Now a temporary instance of std::string has been created that will be deleted at the end of the method call. Thus it's obviously not a good idea to reference an instance that won't exist anymore after the method call. I'd suggest you either change the return type to const string or store the "" in a member or static variable of SomeClass.

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