C++11 std::size() Equivalent Function

c++c++11

Given a builtin array x of arbitrary type T, there are functions std::begin() and std::end() that I can call, but why isn't there a std::size()? Seems odd not to have that.

I could use std::end(x)-std::begin(x), but still a std::size(x) would be better.

Yes, I know of the std::vector and std::array classes. This is just a question of why something as simple as this isn't available as yet in the STL.

Best Answer

Just a note to let people know that N4280 "Non-member size() and more (Revision 2)" has been accepted into the C++17 Working Draft. This includes std::size() as well as std::empty() and std::data().

Related Solutions

C++ – Getting the Size of Member Variable

You can do that in C++0x:

sizeof(foo::a);

C++ – Does C++ Have a Free Function size(object)?

size is actually C++17 functionality. The real benefit to is akin to the benefit of begin and end from C++11.

Note that the first definition of size simply returns the container's size method.

So if I have a templated function like this:

template <typename T>
auto foo(const T& bar) { return bar.size(); }

This could only be used with containers, but if I change that to:

template <typename T>
auto foo(const T& bar) { return size(bar); }

It can be used with C-style arrays too. I've added a live example here: http://melpon.org/wandbox/permlink/Rlpi5wueA14JOW2P

In summary, you should always use size and other range based functions because of the improvements to generality and container agnostic code (see here for more).

Best Answer

Related Solutions

C++ – Getting the Size of Member Variable

C++ – Does C++ Have a Free Function size(object)?

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