std::string array[] = { "one", "two", "three" };
How do I find out the length of the array in code?
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4 Answers
You can use std::begin and std::end, if you have C++11 support.:
int len = std::end(array)-std::begin(array);
// or std::distance(std::begin(array, std::end(array));
Alternatively, you write your own template function:
template< class T, size_t N >
size_t size( const T (&)[N] )
{
return N;
}
size_t len = size(array);
This would work in C++03. If you were to use it in C++11, it would be worth making it a constexpr.
7 Comments
Joseph Mansfield
Or
std::extent<decltype(array)>::value.
juanchopanza
@sftrabbit I hadn't even thought of that one. Nice!
Daniel Frey
@sftrabbit Make that an answer and get my +1
Daniel Frey
Your
len is missing a constexpr :)juanchopanza
@DanielFrey good point, but I wanted to provide a C++03 option. I will add a note about that.
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Use the sizeof()-operator like in
int size = sizeof(array) / sizeof(array[0]);
or better, use std::vector because it offers std::vector::size().
int myints[] = {16,2,77,29};
std::vector<int> fifth (myints, myints + sizeof(myints) / sizeof(int) );
Here is the doc. Consider the range-based example.
3 Comments
NPS
Is it possible to define an
std::vector with initializaiton (like in my example with ordinary array)?
bash.d
C++11 offeres initializers. Otherwise you can create an array as you did and assign it to the std::vector. Ill post the doc in a minute
Cheers and hth. - Alf
This C level expression should not be recommended without explaining its type unsafety and how that can be remedied.
C++11 provides std::extent which gives you the number of elements along the Nth dimension of an array. By default, N is 0, so it gives you the length of the array:
std::extent<decltype(array)>::value
3 Comments
NPS
Does it do all the calculations at compile time (i.e. after compilation it's just a number, needing no extra calculations)?
Cheers and hth. - Alf
@NPS: yes, you can see that from the fact that it's a type
NPS
Any way to make it short (like macro but I'd rather not use macros)?
Like this:
int size = sizeof(array)/sizeof(array[0])
3 Comments
NPS
I knew this one but I thought it might not work with
std::string (due to variable text lengths).Cheers and hth. - Alf
This C level expression should not be recommended without explaining its type unsafety and how that can be remedied.
M.M
@NPS
sizeof(std::string) is always the same ; it does not have anything to do with how many characters are stored in the string