In C++ books, array bound must be constant expression, but why the following code works?

Question

#include <iostream>
using namespace std;

int main(){
    int n=10;
    int a[n];

    for (int i=0; i<n; i++) {
        a[i]=i+1;
        cout<<a[i]<<endl;
}
    return 0;
}

worked fine in Xcode4 under Mac

as said in books, it should be wrong, why?

so confused~

`alloca()` is not standard C++ either. The standard method is in fact `std::vector a(n)` — MSalters, Mar 15 '13 at 12:01
That it "works" doesn't mean it's not wrong :) `char* ptr = "lol"; *ptr = 'a';` sometimes "works" too. — Lightness Races in Orbit, Apr 05 '13 at 12:50
@LightnessRacesinOrbit that reminded me of how I learned not to modify const strings in a very very very painful way :) — holgac, May 02 '13 at 12:23
For much more details also see [Does “int size = 10;” yield a constant expression?](http://stackoverflow.com/q/21273829/1708801) — Shafik Yaghmour, Mar 16 '17 at 17:46

score 6 · Answer 1 · answered May 10 '11 at 09:03

6

This a a C99 feature called VLA which some compilers also allow in C++. It's allocation on stack, just as it would be with int a[10].

answered May 10 '11 at 09:03

Erik

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Nawaz · Answer 2 · 2011-05-10T09:32:28.813

5

That is C99 feature that allows VLA (variable length array).

Compile it with g++ -pedantic, I'm sure that wouldn't compile.

edited May 10 '11 at 09:32

answered May 10 '11 at 09:04

Nawaz

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1

You might also need `-std=c++98` or `-ansi`. – James Kanze May 10 '11 at 09:10

In C++ books, array bound must be constant expression, but why the following code works?

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