How to use the the value stored in a int variable as the size of my array
eg.
int a = 40;
int b[a]; // a = 40
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463035818_is_not_an_ai
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Michael
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This is not valid in `c++` https://stackoverflow.com/questions/1887097/why-arent-variable-length-arrays-part-of-the-c-standard – drescherjm May 20 '19 at 11:14
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[Variable-length arrays](https://en.wikipedia.org/wiki/Variable-length_array) are not supported in C++ (though some compilers add it as a non-portable extension). Use [`std::vector`](https://en.cppreference.com/w/cpp/container/vector) instead. – Some programmer dude May 20 '19 at 11:14
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2That has been asked and answered many many many times. – Aykhan Hagverdili May 20 '19 at 11:23
2 Answers
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You can't; not in standard C++ anyway, unless a is a constexpr or const integral type with a positive value.
The best alternative is a std::vector<int> b(a);
Bathsheba
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2`constexpr` is not necessary for `a` to be a constant expression. `const int a = 40;` would be sufficient in this case. That said, I don't see a reason not to use `constexpr`. – eerorika May 20 '19 at 11:28
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Create an array with new[]:
int *b = new int[a];
but remember to delete it later with:
delete[] b;
A better alternative is the std container, like std::vector.
