Differentiate between Array and Pointer in function parameter - C++

Question

This is a problem I've been struggling with for a while now. The codes I have tried below are not compiling either.

The question is: How do I differentiate between a pointer and a fixed array in a function parameter?

// Concepts for the Array I have tried but not succeeded.
template <size_t length, typename type>
const unsigned int len(type arg[static length]) { return length; }

template <size_t length, typename type>
const unsigned int len(type(&)[length]) { return length; }

// This works for Arrays & Pointers
// but should be less prioritized over the function that detects the Array
template <typename type> const unsigned int len(type* arg) { return sizeof(*arg); }

I know how arrays & pointers are basically similar when passed into a function, but it asks the question: is there not a way to tell them apart?

Syntactically yes, but otherwise what other way is there?

Anyway, thanks for reading through and cheers on your response.

static length is not valid C++. That particular C99 feature is not available. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Nov 21, 2018 at 6:30

Jeremy Friesner · Accepted Answer · 2018-11-21 06:30:30Z

1

This approach works for me:

#include <stdio.h>

template<typename T, int size> unsigned int len(const T(&)[size]) {printf("The number of items in your array is:  %i\n", size); return size;}
template<typename T> unsigned int len(const T * p) {printf("The size of the item your pointer points to is: %zu\n", sizeof(*p)); return sizeof(*p);}

int main(int, char **)
{
   int myArray[10];
   int * myPointer = myArray;

   (void) len(myArray);
   (void) len(myPointer);

   return 0;
}

... when I run it, it prints out:

The number of items in your array is:  10
The size of the item your pointer points to is: 4

answered Nov 21, 2018 at 6:30

Jeremy Friesner

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Comments

Toby Speight · Accepted Answer · 2018-11-21 09:43:04Z

The length of an array can only be deduced if the array is passed by reference, otherwise it decays to pointer:

template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
//                               ^^^

template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
//                                ^^^

Full demo:

#include <cstdlib>

template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }

template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }

#include <iostream>
int main(int, char **)
{
    const int array[7] = {};
    const int *pointer = array;

    std::cout << "array has " << len(array) << " items\n"
              << "and pointer is to " << len(pointer) << " chars\n";
}

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