jackhab jackhab - 1 month ago 5
C++ Question

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.

Answer

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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