CODEWITHSUNDEEP
c++referencereturn-valuervaluetemporary
Gaurav
Gaurav

Reputation: 201

Warning: returning reference to temporary - strange case (Clarification for Rvalue)

In this code:

const int & fun(const int &i)
{
    return 2*i;
}

int main()
{
    const int k=3;
    cout<<fun(k)<<endl;
    return 0;
}

In this case the parameters of fun are not local (No Temporary object made to store the Reference return type) then why this warning? + If I remove const from the return type of function fun it says

error: invalid initialization of non-const reference of type ‘int&’ from an rvalue of type ‘int’

But on removing 2* (only i left as value being returned) it doesn't show any error -> What I can think is this 2* thing is converting the return into Rvalue but then isn't the return value an Rvalue itself? Where am I going wrong?

Upvotes: 1

Views: 620

Answers (1)

NathanOliver
NathanOliver

Reputation: 180500

return 2*i; does not multiply 2 by i and store that result into i. It multiples 2 by i and puts that result into a temporary. Trying to return that temporary by reference is a no go as it is destroyed at the end of that line. lifetime extension does not apply here.

If you want to modify i and return it you need to use operator *= which will modify i and give you a reference to it that you can return like

return i *= 2;

But fun would need to take a int& instead of const int &.

If you want to return an rvalue then what you do is return by value like:

int fun(const int &i)
{
    return 2*i;
}

and now you can capture it like:

int main()
{
    const int& ret = fun(3);
    cout << ret << endl;
    return 0;
}

Upvotes: 5

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