Compile time sizeof_array without using a macro - c++

Question

This is just something that has bothered me for the last couple of days, I don't think it's possible to solve but I've seen template magic before.
Here goes:
To get the number of elements in a standard C++ array I could use either a macro (1), or a typesafe inline function (2):
(1)
#define sizeof_array(ARRAY) (sizeof(ARRAY)/sizeof(ARRAY[0]))
(2)
template <typename T>
size_t sizeof_array(const T& ARRAY){
return (sizeof(ARRAY)/sizeof(ARRAY[0]));
}
As you can see, the first one has the problem of being a macro (for the moment I consider that a problem) and the other one has the problem of not being able to get the size of an array at compile time; ie I can't write:
enum ENUM{N=sizeof_array(ARRAY)};
or
BOOST_STATIC_ASSERT(sizeof_array(ARRAY)==10);// Assuming the size 10..
Does anyone know if this can be solved?
Update:
This question was created before constexpr was introduced. Nowadays you can simply use:
template <typename T>
constexpr auto sizeof_array(const T& iarray) {
return (sizeof(iarray) / sizeof(iarray[0]));
}

Answer 1

Try the following from here:
template <typename T, size_t N>
char ( &_ArraySizeHelper( T (&array)[N] ))[N];
#define mycountof( array ) (sizeof( _ArraySizeHelper( array ) ))
int testarray[10];
enum { testsize = mycountof(testarray) };
void test() {
printf("The array count is: %d\n", testsize);
}
It should print out: "The array count is: 10"

Answer 2

In C++1x constexpr will get you that:
template <typename T, size_t N>
constexpr size_t countof(T(&)[N])
{
return N;
}

Answer 3

The best I can think of is this:
template <class T, std::size_t N>
char (&sizeof_array(T (&a)[N]))[N];
// As litb noted in comments, you need this overload to handle array rvalues
// correctly (e.g. when array is a member of a struct returned from function),
// since they won't bind to non-const reference in the overload above.
template <class T, std::size_t N>
char (&sizeof_array(const T (&a)[N]))[N];
which has to be used with another sizeof:
int main()
{
int a[10];
int n = sizeof(sizeof_array(a));
std::cout << n << std::endl;
}
[EDIT]
Come to think of it, I believe this is provably impossible to do in a single "function-like call" in C++03, apart from macros, and here's why.
On one hand, you will clearly need template parameter deduction to obtain size of array (either directly, or via sizeof as you do). However, template parameter deduction is only applicable to functions, and not to classes; i.e. you can have a template parameter R of type reference-to-array-of-N, where N is another template parameter, but you'll have to provide both R and N at the point of the call; if you want to deduce N from R, only a function call can do that.
On the other hand, the only way any expression involving a function call can be constant is when it's inside sizeof. Anything else (e.g. accessing a static or enum member on return value of function) still requires the function call to occur, which obviously means this won't be a constant expression.

Answer 4

It's not exactly what you're looking for, but it's close - a snippet from winnt.h which includes some explanation of what the #$%^ it's doing:
//
// RtlpNumberOf is a function that takes a reference to an array of N Ts.
//
// typedef T array_of_T[N];
// typedef array_of_T &reference_to_array_of_T;
//
// RtlpNumberOf returns a pointer to an array of N chars.
// We could return a reference instead of a pointer but older compilers do not accept that.
//
// typedef char array_of_char[N];
// typedef array_of_char *pointer_to_array_of_char;
//
// sizeof(array_of_char) == N
// sizeof(*pointer_to_array_of_char) == N
//
// pointer_to_array_of_char RtlpNumberOf(reference_to_array_of_T);
//
// We never even call RtlpNumberOf, we just take the size of dereferencing its return type.
// We do not even implement RtlpNumberOf, we just decare it.
//
// Attempts to pass pointers instead of arrays to this macro result in compile time errors.
// That is the point.
//
extern "C++" // templates cannot be declared to have 'C' linkage
template <typename T, size_t N>
char (*RtlpNumberOf( UNALIGNED T (&)[N] ))[N];
#define RTL_NUMBER_OF_V2(A) (sizeof(*RtlpNumberOf(A)))
The RTL_NUMBER_OF_V2() macro ends up being used in the more readable ARRAYSIZE() macro.
Matthew Wilson's "Imperfect C++" book also has a discussion of the techniques that are used here.

Answer 5

The Problem
I like Adisak's answer:
template <typename T, size_t N>
char ( &_ArraySizeHelper( T (&arr)[N] ))[N];
#define COUNTOF( arr ) (sizeof( _ArraySizeHelper( arr ) ))
It's what Microsoft uses for the _countof macro in VS2008, and it's got some nice features:
It operates at compile time
It's typesafe (i.e. it will generate a compile-time error if you give it a pointer, which arrays degrade too all too easily)
But as pointed out by Georg, this approach uses templates, so it's not guaranteed to work with local types for C++03:
void i_am_a_banana() {
struct { int i; } arr[10];
std::cout << COUNTOF(arr) << std::endl; // forbidden in C++03
}
Fortunately, we're not out luck.
The Solution
Ivan Johnson came up with a clever approach that wins on all accounts: it's typesafe, compile-time, and works with local types:
#define COUNTOF(arr) ( \
0 * sizeof(reinterpret_cast<const ::Bad_arg_to_COUNTOF*>(arr)) + \
0 * sizeof(::Bad_arg_to_COUNTOF::check_type((arr), &(arr))) + \
sizeof(arr) / sizeof((arr)[0]) )
struct Bad_arg_to_COUNTOF {
class Is_pointer; // incomplete
class Is_array {};
template <typename T>
static Is_pointer check_type(const T*, const T* const*);
static Is_array check_type(const void*, const void*);
};
For those who are interested, it works by inserting two "tests" before the standard sizeof-based array-size macro. Those tests don't impact the final calculation, but are designed to generate compile errors for non-array types:
The first test fails unless arr is integral, enum, pointer, or array. reinterpret_cast<const T*> should fail for any other types.
The second test fails for integral, enum, or pointer types.
Integral and enum types will fail because there's no version of check_type that they match, since check_type expects pointers.
Pointer types will fail because they'll match the templated version of check_type, but the return type (Is_pointer) for the templated check_type is incomplete, which will produce an error.
Array types will pass because taking the address of an array of type T
will give you T (*)[], aka a pointer-to-an-array, not a pointer-to-a-pointer. That means that the templated version of check_type won't match. Thanks to SFINAE, the compiler will move on to the non-templated version of check_type, which should accept any pair of pointers. Since the return type for the non-templated version is defined completely, no error will be produced. And since we're not dealing with templates now, local types work fine.

Answer 6

If you are on a Microsoft only platform, you can take advantage of the _countof macro. This is a non-standard extension which will return the count of elements within an array. It's advantage over most countof style macros is that it will cause a compilation error if it's used on a non-array type.
The following works just fine (VS 2008 RTM)
static int ARRAY[5];
enum ENUM{N=_countof(ARRAY)};
But once again, it's MS specific so this may not work for you.

Answer 7

You can't solve it in general, thats one the reasons for array wrappers like boost array (plus stl-style behaviour of course).

Answer 8

It appears not to be possible to obtain the sizeof array as a compile-time constant without a macro with current C++ standard (you need a function to deduce the array size, but function calls are not allowed where you need a compile-time constant). [Edit: But see Minaev's brilliant solution!]
However, your template version isn't typesafe either and suffers from the same problem as the macro: it also accepts pointers and notably arrays decayed to a pointer. When it accepts a pointer, the result of sizeof(T*) / sizeof(T) cannot be meaningful.
Better:
template <typename T, size_t N>
size_t sizeof_array(T (&)[N]){
return N;
}

Answer 9

Without C++0x, the closest I can get is:
#include <iostream>
template <typename T>
struct count_of_type
{
};
template <typename T, unsigned N>
struct count_of_type<T[N]>
{
enum { value = N };
};
template <typename T, unsigned N>
unsigned count_of ( const T (&) [N] )
{
return N;
};
int main ()
{
std::cout << count_of_type<int[20]>::value << std::endl;
std::cout << count_of_type<char[42]>::value << std::endl;
// std::cout << count_of_type<char*>::value << std::endl; // compile error
int foo[1234];
std::cout << count_of(foo) << std::endl;
const char* bar = "wibble";
// std::cout << count_of( bar ) << std::endl; // compile error
enum E1 { N = count_of_type<int[1234]>::value } ;
return 0;
}
which either gives you a function you can pass the variable to, or a template you can pass the type too. You can't use the function for a compile time constant, but most cases you know the type, even if only as template parameter.

Answer 10

Now STL libraries are available to decide/select array size compile time
#include <iostream>
#include <array>
template<class T>
void test(T t)
{
int a[std::tuple_size<T>::value]; // can be used at compile time
std::cout << std::tuple_size<T>::value << '\n';
}
int main()
{
std::array<float, 3> arr;
test(arr);
}
Output:
3