Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
935 views
in Technique[技术] by (71.8m points)

c++ - warning: returning reference to temporary

I have a function like this

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

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.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...