C++- "using" keyword or using namespace name with scope resolution operator

Question

various programs in C++ are written without using scope resolution operator in this way:

#include <iostream>
#include <string>

int main()
{
std::string name = "My Name";
std::cout << name << std::endl;

return 0;
}

and I have also seen using "using" keyword:

#include <iostream>
#include <string>
using namespace std;
int main()
{
string name = "My Name";
cout << name << endl;

return 0;
}

which one is better for efficiency reasons?

What do you mean _"without using scope resolution operator"_? Your example uses it, twice. Do you mean _with_? — Jonathan Wakely, Dec 22 '16 at 10:41

Dietmar Kühl · Answer 1 · 2016-12-22T10:30:37.027

Assuming the same names end up being found the efficiency is the same: independent of how the function name is spelled the same function is being called. The main difference is how names are being located. Using full qualification prevents, e.g., argument-dependent look-up and is, thus, easier to understand.

Of course, when you actually do have a customization point you'd want to use an unqualified call for argument-dependent look-up to kick in. If there is no need for a default implementation for the customization point there is no need to have a using declaration or even a using directive. Incorrect use of customization points can have a negative performance impact. For example, when using swap() you do not want use the name qualified:

template <typename T>
void some_function(T& v1, T& v2) {
    std::swap(v1, v2); // <--- this is bad! It uses the default implementation

    using std::swap;   // make a default implementation visible
    swap(v1, v2);      // <--- this is better: if it exists uses T's swap()
}

If T has a customized swap() it is likely more efficient than std::swap() which may cause copies of the values v1 and v2. With move-enabled types T the difference isn't as bad but it could still be substantial. Of course, the issue here isn't the use of qualification or no qualification but rather the fact that the two ways to call a function may result in different functions being found depending on whether swap() is overloaded for T.

BTW, if you are interested in efficiency, do not use std::endl!

@ProthomOntim: yes - that doesn't flush the stream all the time. — Dietmar Kühl, Dec 22 '16 at 10:24
@T.C.: good point. I have changed the code snippet to not use `std::vector`. — Dietmar Kühl, Dec 22 '16 at 10:31

score 1 · Answer 2 · answered Dec 22 '16 at 10:53

The two programs should produce identical executable code. In C++ name lookup happens at compile-time, not run-time, so the compiler finds the name during compilation and then nothing changes later. There is no run-time evaluation, so the syntax used to find a given name can't affect the program's efficiency.

As Dietmar says, the choice of syntax might cause a different name to be found, which could alter the behaviour, but assuming the same name is found there can be no difference in efficiency.

C++- "using" keyword or using namespace name with scope resolution operator

2 Answers2