How can I create an array whose length is specified by a const argument?

Question

I have a function:

void f(const size_t &len){
    double arr[len];
}

but it doesn't work, because I got message from len, that "expression must have a constant value". How can I solve this problem, if I don't want to create a global variable len?

`double* arr = new double[len];`? Or, even better, `std::vector arr(len);`? — Daniel Langr, Feb 20 '20 at 10:38
`const` and constant expression are different... Former is more "read-only" whereas later is "value known at compile time". — Jarod42, Feb 20 '20 at 10:38
Your code doesn't make much sense. Why do you want to do that ? Why not a `std::vector` ? — kebs, Feb 20 '20 at 10:38
new() is slow for my big case, what about vector, Yes, maybe it was obviously, sorry — NN_05, Feb 20 '20 at 10:41
@KK_1999 Vectors just use `new` internally (allocate memory). — Daniel Langr, Feb 20 '20 at 10:42
By the way what difference between vector and usual array in C++? Can usual array be faster than std::vector? — NN_05, Feb 20 '20 at 10:51
@KK_1999 Memory layout of elements is the very same (contiguous storage). From this perspective, there are no differences. — Daniel Langr, Feb 20 '20 at 10:54
@KK_1999 Note that an interesting trade-off solution might be for you a vector with _small data/buffer optimization_, such as `boost::container::small_vector`. But it depends on the values of `len` in your code runs. — Daniel Langr, Feb 20 '20 at 11:00
@DanielLangr and `sizeof(T)`. E.g. never really applicable to `std::vector` — Caleth, Feb 20 '20 at 13:15
I tried to use vector but unfortunately it was slow solution for me — NN_05, Feb 20 '20 at 17:21

divinas · Answer 1 · 2020-02-20T13:11:13.410

C++ does not support variable-length arrays. If len is a compile time constant, I'd advise you to use std::array with a template, like so:

template<size_t len>
void f(){
    std::array<double, len> arr;
    //use arr
}

And you would use it like that:

int main()
{
    f<5>();
}

Do note, that the 5 in my example is a compile-time constant. If you do not know the size of your array at compile time, use a std::vector. You'd do this like this:

void f(const size_t len){
    std::vector<double> arr(len);
    //use arr
}

int main()
{
    size_t variableLength = 0;
    std::cin >> variableLength;
    f(variableLenght);
}

score 0 · Answer 2 · answered Feb 20 '20 at 10:48

0

You need to create a Dynamic Array (On the free store). using one of the memory allocation functions.

double* arr = new double[len];

But you have to keep track of the memory release yourself!! so you have to Save your pointer as a global variable or as a class member to delete it later using

delete[] arr;
arr = nullptr;

This is how Dynamic arrays are managed in c++, but static arrays are created on the stack of an instruction block and they are destroyed automatically.And theire size must be defined at compile time not at run time.

answered Feb 20 '20 at 10:48

Rahim

1
3

_But you have to keep track of the memory release yourself!_ — That's why it's so much preferred to use a `vector`. – Daniel Langr Feb 20 '20 at 10:56
@ Daniel Langr. Yes you are right but is very important and benefic to be familiar with memory managment. because this is the strong point in a Low level programing Language. Otherwise i would go for something else like Java or C# where all this stuff is hidden from the programmer – Rahim Feb 20 '20 at 11:00
_Otherwise i would go for something else like Java or C#..._ — Agree, but one isn't always so lucky to have this choice ;-). – Daniel Langr Feb 20 '20 at 11:06

How can I create an array whose length is specified by a const argument?

2 Answers2