CODEWITHSUNDEEP
c++templatesstdarray
Dean
Dean

Reputation: 6950

Reading range from std::array

How can I accept a std::array which might have a different dimension? This should be known at compile-time but the following won't work

template<int n>
void read_interval(size_t start, size_t end, std::array<n, char>& dest)

I also know that end-start == n so that might be somehow templated either.

Upvotes: 0

Views: 117

Answers (3)

fghj
fghj

Reputation: 9394

Such code compiles, you should use size_t instead of int as template parameter

#include <array>

template<size_t n>
void read_interval(size_t start, size_t end, std::array<char, n>& dest)
{
}

int main()
{
    std::array<char, 10> arr1;

    read_interval(0, 10, arr1);

    std::array<char, 8> arr2;

    read_interval(0, 8, arr2);
}

end if n is always equal end you can just use n inside read_interval as ordinary constant.

Upvotes: 0

eerorika
eerorika

Reputation: 238341

You need to template the size argument, just like you did. Except you've swapped the order of the template parameters of std::array which is why it doesn't work.

Upvotes: 0

TartanLlama
TartanLlama

Reputation: 65620

You have the template arguments for std::array the wrong way round and the non-type argument is a std::size_t, not an int:

template<std::size_t n>
void read_interval(size_t start, size_t end, std::array<char,n>& dest)
{
    //...
}

You can't statically ensure that end - start == n as start and end are runtime values. If you really need that static assurance, you'll need to make them template parameters, otherwise you can use a runtime assert for debug mode or carry out a check and throw an exception.

Upvotes: 1

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