c++ extern constant int for array size - c++

Question

I have the following three files in my code (with most of the code removed. This is just to isolate the issue).
global.h:
//global.h
#ifndef GLOBAL_H
#define GLOBAL_H
extern const int ARRAYSIZEX;
extern const int ARRAYSIZEY;
extern const int ARRAYSIZEZ;
#endif //GLOBAL_H
global.cpp:
//global.cpp
#include "global.h"
const int ARRAYSIZEX = 5;
const int ARRAYSIZEY = 2;
const int ARRAYSIZEZ = 4;
main:
//main
#include "global.h"
using namespace std;
someType mySomeTypeArray[ARRAYSIZEX][ARRAYSIZEY][ARRAYSIZEZ];
int main(int argc, char **argv)
{
//...
}
Compiling gives me three errors at the declaration of mySomeTypeArray.
error: array bound is not an integer constant before ']' token
I want to keep my global variable and array size definitions in global.h/cpp for this application, just for organization, so that all my configuration parameters are in one place. What's the proper way to achieve what I'm trying to do?
Thanks

Answer 1

Your declaration is failing because array sizes need to be evaluated at compile-time and your encapsulation scheme is actually hiding the values from the compiler. This is true because compilers work on individual translation units. While compiling main.cpp your compiler sees only extern const int ARRAYSIZEX thanks to the include statement, but not the value which is visible in a separate translation unit so it can't figure out the memory layout.
While const variables can used as array sizes in some contexts, the language provides the more appropriate constexpr qualifier which comes with a set of restrictions that enforce its compile-time evaluation and suitability for array sizes. I recommend always using it when appropriate because it will point you to the error in situations such as this. In this case, you would get a compiler error because an extern constexpr declaration is ill-formed which hints at the proper solution: to hold the values for compile-time constants directly inside the header file.
global.h
constexpr int ARRAYSIZEX = ...;
constexpr int ARRAYSIZEY = ...;
constexpr int ARRAYSIZEZ = ...;
main.cpp
#include "global.h"
someType mySomeTypeArray[ARRAYSIZEX][ARRAYSIZEY][ARRAYSIZEZ];

Answer 2

The problem here is extern int x means "x is defined in another file, but don't worry about the particulars, all you need to know is it's an int". This is normally good enough, except when the compiler needs to know right there and then what x is.
Since that's defined in a whole other file it can't. That file must be compiled before it knows, and the result of that compilation, due to the way C++ works, can't impact the compilation of this file.
You'll need to declare that as a const int in a header if you want to share those values. extern int won't cut it.
Although this is a trivial example, there's really no reason to go down the extern road at all. Just define the values in the header file as regular const int.

Answer 3

Array size must be specified by an integer constant expression. A const int object can be used in an integer constant expression if and only if it declared with an initializer and that initializer is also an integer constant expression. Your ARRAYSIZE... variables do not satisfy that requirement. In main they are declared without an initializer. You cannot use ARRAYSIZE... variables as array sizes in main.
Unless you have a specific requirement to give these variables external linkage, simply declare (and define) them in the header as
const int ARRAYSIZEX = 5;
const int ARRAYSIZEY = 2;
const int ARRAYSIZEZ = 4;
These object will have internal linkage though, which is different from what your original variant attempts to do.
If really want to give them external linkage, declare them as inline extern const in the header
inline extern const int ARRAYSIZEX = 5;
inline extern const int ARRAYSIZEY = 2;
inline extern const int ARRAYSIZEZ = 4;
Since inline by itself prevents const from imposing internal linkage, extern is entirely optional in these declarations. And since inline const combination can be replaced with constexpr (as #M.M noted in the comments), you can achieve the same effect with just
constexpr int ARRAYSIZEX = 5;
constexpr int ARRAYSIZEY = 2;
constexpr int ARRAYSIZEZ = 4;

Answer 4

The problem here is that ARRAYSIZEX, ARRAYSIZEY and ARRAYSIZEZ are not the compile time constants. They are constants - so their values can't be changed but their values are not known to compiler.
In C++ the compilation process consists of 3 basic steps.
Preprocessing of all source files done by preprocessor.
Compilation for every translation unit (.cpp file) done by compiler. For every translation unit compiler creates an object file.
Linking of all object files done by linker. The output is an executable file.
In C++ the keyword extern for compiler means that the variable is 'somewhere' defined. The compiler doesn't know the variable's real address but by placing keyword extern it's assured that the variable really exists and the linker will be able to find its address by its name when creating the executable file.
The problem here is that compiler in step 2 wants to create the object file but it doesn't known how big the array is going to be because it doesn't know the value of these constants. Yes, linker in step 3 will finally find them when putting together all object files but it's too late for compiler. So it generates that error.
The solution is simple. Use already mentioned constexpr keyword and initialize all variables with initializers. The keyword constexpr marks compile time constants - constants that have to be initialized in initializers and are known to compiler.