Passing to a Reference Argument by Value

Question

Consider this simple program:

vector<int> foo = {0, 42, 0, 42, 0, 42};
replace(begin(foo), end(foo), foo.front(), 13);

for(const auto& i : foo) cout << i << '\t';

When I wrote it I expected to get:

13 42 13 42 13 42

But instead I got:

13 42 0 42 0 42

The problem of course is that replace takes in the last 2 parameters by reference. So if either of them happen to be in the range being operated on the results may be unexpected. I can solve this by adding a temporary variable:

vector<int> foo = {0, 42, 0, 42, 0, 42};
const auto temp = foo.front();
replace(begin(foo), end(foo), temp, 13);

for(const auto& i : foo) cout << i << '\t';

I do know that C++11 gave us all kinds of type tools is it possible that I could simply force this value to a non-reference type and pass that inline, without creating the temporary?

Answer 1

A solution could be as follows (even though you are making a temporary)

template<class T>
void replace_value_of_first(std::vector<T>& v, const T& value)
{
    std::replace(v.begin(), v.end(), T(v.front()), value);
}

Answer 2

You can write a simple function that takes in a reference and returns a value. this will "convert" the reference into a value. This does generate a temporary but it is unnamed and will be destroyed at the end of the full expression. Something like

template<typename T>
T value(const T& ref)
{
    return ref;
}

And then you can use it like

int main()                                                   
{                                                            
    vector<int> foo = {0, 42, 0, 42, 0, 42};
    replace(begin(foo), end(foo), value(foo.front()), 13);

    for(const auto& i : foo) cout << i << '\t';                                      
}

output:

13  42  13  42  13  42

Live Example