Define an array with lengh equal to vector::size()

Question

I am surprised that below codes compile and display a right result.

#include <iostream>
#include <vector>

int main()
{
    std::vector<int> v;
    v.push_back(1);
    v.push_back(2);
    v.push_back(3);

    char arr[v.size()];  /* why does this line compile? */

    std::cout << sizeof(arr) / sizeof(arr[0]) << std::endl;  /* display 3 */

    return 0;
}

I believe a constant length must be used when defining an array in C/C++. The vector::size() is determined in runtime, so why does this work?

Jean-Baptiste Yunès · Answer 1 · 2018-08-12T07:28:38.540

2

First, C/C++ are two different languages.

C99 defined VLAs (Variable Length Array), but this has been changed to optional in subsequent standards.

C++ doesn't define VLAs.

Thus, this compiles because you are using a compiler that has an extension for, probably GCC. Then strictly speaking, your code is not C++ standard compliant.

edited Aug 12 '18 at 07:28

answered Aug 12 '18 at 07:20

Jean-Baptiste Yunès

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It became optional in C11 rather than removed. – user7860670 Aug 12 '18 at 07:24
@VTT Many thanks, changed – Jean-Baptiste Yunès Aug 12 '18 at 07:29
@vtt Yeah, it's about time this feature became mandated by the standard imo – Paul Sanders Aug 12 '18 at 07:34
1

@Jean-BaptisteYunès Thanks! I wasn't aware that even I had used a normal int variable there, it would still compile in C++. After adding -Wvla, it shows warnings. – Chris XJ Zhu Aug 12 '18 at 07:46

Define an array with lengh equal to vector::size()

1 Answers1