returning a string literal to const String reference in C++

Question

When compiling the following code the compiler issue the warning warning: returning reference to temporary

const string& example1()
{
    return "Hello";
}

This code does not even compile:

void example2(){
    const string& str = "Hello";
}

and this one is valid since we know that a literal string is initialized into read-only memory segment by the compiler.

char* example3()
{
    return "Hello";
}

Could you please help me understand what happen behind the scene when compiling the method example1() ?

Thank you very much for your help !

Since the method return an std::string, a temporary std::string is creating from the string literal. This temporary is destroyed at the end of the function scope. — quantdev
– quantdev, Commented Feb 1, 2015 at 4:34

Jérôme · Accepted Answer · 2015-02-01 04:42:32Z

4

Return "Hello" creates a temporary std::string which will be deleted at the end of your function. Here, you are returning a reference on a std::string that won't exist at the end of the function call.

To solve that you can change the return type of example1() to string

edited Feb 1, 2015 at 4:42

answered Feb 1, 2015 at 4:37

Jérôme

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2 Comments

flogram_dev Over a year ago

Preferably just string without const.

Tomer G Over a year ago

Thank you very much Jerome and Zenith. But how come then a temporary string is not created in the case of example2() ?

flogram_dev · Accepted Answer · 2015-02-01 05:02:53Z

2

Adding to Jerome's answer, you can also do this if you only want to have a single copy of the string:

const string& example1()
{
    static string example = "Hello";
    return example;
}

(Now example is a static function variable, and will also exist outside the function.)

answered Feb 1, 2015 at 5:02

flogram_dev

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