[RESOLVED] I do not why this compiles! confused

[ar115]

Hi,
I am sure that the size of c style arrays must be known at compile tim
but the code below compiles. I am using g++ 4.2.4 and really confused!

int main(){
int N = 5;
int x[N];
x[2] = 3;
}

[monarch_dodra]

Hi,
I am sure that the size of c style arrays must be known at compile tim
but the code below compiles. I am using g++ 4.2.4 and really confused!

int main(){
int N = 5;
int x[N];
x[2] = 3;
}

The compiler sees that N is defined by the time you reach int x[N], N's value can only be 5, which means the size is known at compile time. The compiler does not necessarily need the size to be a const.
If you volatile the N, it would not compile.

[ar115]

but it also works without setting the value of N like this

int main(){
int N;
int x[N];
}

[Russco]

AFAIK that should not compile. The size of arrays must be a nonzero constant.

[monarch_dodra]

int main()
{
    int N = 5;
    ++N;
    N*=2;
    int i;
    std::cin >> i;
    if (i==1) ++N;
    int x[N];
    x[2] = 3;
    std::cout << N << std::endl;
    std::cout << sizeof(x) << std::endl;
}

Now I'm confused too, as this example clearly shows that the size of x is not only unkown at compile time, it is correctly (and dynamically) sized to 12*4 or 13*4 according to i.

[Philip Nicoletti]

It is a g++ extension .. to turn off, compile with:

g++ -pedantic