Hello, is this thing on?

[jeffkurtz]

#include <iostream>
using namespace std;

void create(const int);

int main()
{
const int s=26;
create(s);
}

void create(const int s)
{
const int a=26;
int c[s];
}

The above program will not compile and shows "error c2057: expected constant expression." However, variables s clearly is a constant expression. If I change the "s" to "a" in "int c[s]" the program runs fine. This makes absolutely no sense- both variables are of the same type and are both constant, but the program will compile with one and not the other. The software environment is MVS 2005, but I've tried others with the same results. Can anyone tell me why? Thanks, Jeff Kurtz

[valaris]

Because the compiler will know how big the stack frame will need to be when you use a, but s could be anything. If you need to dynamically allocate some memory use the heap.

[Mr.Pointer]

I used g++ for the above code and it compiled.

Why are you declaring an array with a variable size? Not all compilers can support it.

EDIT: too late :P

[cpjust]

Variable sized arrays is a C99 feature if I'm not mistaken, and VC++ doesn't support C99.

[valaris]

Variable sized arrays is a C99 feature if I'm not mistaken, and VC++ doesn't support C99.

I didn't know that "variable" sized arrays on the stack could exist? How does this work under the hood?

[matsp]

I didn't know that "variable" sized arrays on the stack could exist? How does this work under the hood?

The compiler produces the relevant calculations to allocate stack-space. Just like for fixed size arrays, but with the difference that the inputs is a variable.

As stated, this is not supported by all compilers.

--
Mats