"built-in" functions in C++

Question

I'm beginner at C++ so if the answer is obvious it possibly is the one I'm looking for. I was reading the second response in this thread and got confused.

#include <algorithm>
#include <cassert>

int
main()
{
    using std::swap;
    int a(3), b(5);
    swap(a, b);
    assert(a == 5 && b == 3);
}

What I don't get is "This is just a defined function. What I meant was, why it is not directly built-in" but there was no need to include a new library so isn't it built-in? Does the std library automatically get imported (if yes, why doesn't the namespace automatically get set to std)?

Solution 1

"This is just a defined function. What I meant was, why it is not directly built-in" but there was no need to import a new library so isn't it built-in? Does the std library automatically get imported (if yes, why doesn't the namespace automatically get set to std)?

Well, by defined function it means, most likely that the function is already pre-written, and defined in the library, it isn't directly built-in probably because it was designed that way; only core essentials were included in the language, and everything else is in a library so the programmer can import what they want.

By built-in, usually it's a keyword, like for or while.

And no, std isn't automatically imported since it's designed so the programmer can choose what namespaces they want, like custom namespaces or std. An example of this being bad to automatically have std is this:

Say you automatically defined std, then you wanted to do using namespace foo;, now if foo also had function cout, you would run into a huge problem, like say you wanted to do this;

// code above
cout << "Hello, World" << endl;
// code below

how would the compiler which namespace function to use use? the default or your foo namespace cout? In order to prevent this, there is no default namespace set, leaving it up to the programmer.

Hope that helps!

Solution 2

This is just a defined function. What I meant was, why it is not directly built-in" but there was no need to include a new library so isn't it built-in?

The C++ library is a part of C++, by definition. However, it is not a part of the core language. C++ is a huge, huge language. Making it so the compiler knew every nook and cranny of the language right off the bat would make the compiler huge and slow to load. The philosophy instead is to keep the core somewhat small and give programmers the ability to extend the functionality by #including header files that. specify what they need.

Why doesn't the namespace automatically get set to std?

That would essentially make all kinds of very common words into keywords. The list of words you shouldn't use (keywords, global function in C, words reserved by POSIX or Microsoft, ...) is already huge. Putting the C++ standard library in namespace std is a feature. Putting all of those names in the global namespace would be a huge misfeature.

Solution 3

In your code, you have the line:

using std::swap;

So the call to swap doesn't need std::. For assert, it was defined as a macro, so it also would not need std::. If you did not use using, then other than macros, you would have to use std:: to refer to functions and objects provided by the standard C++ library.

The standard C++ library normally get linked into your program when you compile your program to create an executable. From this point of view, you might consider it "built-in". However, the term "built-in" would usually mean the compiler treats the word swap like a keyword, which is not the case here. swap is a template function defined in the algorithm header file, and assert is a macro defined in cassert.

A namespace is a convenience to allow parts of software to be easily partitioned by name. So if you wanted to define your own swap function, you could put it into your own namespace.

namespace mine {
    template <typename T> void swap (T &a, T &b) { /*...*/ }
}

And it would not collide with the standard, or with some library that defined swap without a namespace.