C++ - how to pass an array into a function that takes an int * type, within a function

Question

I usually use vectors in C++, but in a particular case I have to use arrays which I'm not used to. If I do this:

// GetArraySize.cpp

#include <iostream>

#include <conio.h>   // remove this line if not using Windows

int main(void)
{
  int myArray[] = { 53, 87, 34, 83, 95, 28, 46 };

  auto arraySize = std::end(myArray) - std::begin(myArray);

  std::cout << "arraySize = " << arraySize << "\n\n";

  _getch();    // remove this line if not using Windows

  return(0);
}

This works as expected (arraySize prints out as 7). But if I do this:

// GetArraySizeWithFunc.cpp

#include <iostream>

#include <conio.h>    // remove this line if not using Windows

// function prototypes
int getArraySize(int intArray[]);

int main(void)
{
  int myArray[] = { 53, 87, 34, 83, 95, 28, 46 };

  int arraySize = getArraySize(myArray);

  std::cout << "arraySize = " << arraySize << "\n\n";

  _getch();    // remove this line if not using Windows

  return(0);
}

int getArraySize(int intArray[])
{
  auto arraySize = std::end(intArray) - std::begin(intArray);

  return((int)arraySize);
}

On the line auto arraySize = std::end(intArray) - std::begin(intArray); I get the error:

no instance of overloaded function "std::end" matches the argument list, argument types are: (int *)

What am I doing wrong?

I should mention a few things:

-I'm aware that with C++ 17 I could use std::size(myArray), but in the context I'm working in I can't use C++ 17

-There may be other / better ways to write a getArraySize() function, but moreover I'm trying to better understand how old-style arrays are passed into / out of functions

"What am I doing wrong?" Well, int main(void) should be int main(). And return(0); should be return 0;. And return((int)arraySize); should be return (int)arraySize;. Instead of a C array, you should use std::array. If you insist on passing in a C array, you should also pass in its length (if its length is something you care about). So int getArraySize(int arraySize) { return arraySize; }, which is a bit redundant. — Eljay
– Eljay, Commented Feb 9, 2019 at 13:22
Don't do an array to pointer decay clang.llvm.org/extra/clang-tidy/checks/…. — Werner Henze
– Werner Henze, Commented Feb 9, 2019 at 13:30
If this is C++, main(void) should be replaced with main(). Also the return type should be int. — HolyBlackCat
– HolyBlackCat, Commented Feb 9, 2019 at 13:42
Also, you should avoid casting by using proper type and if you really need some cat, then you should use C++ cast like static_cast. — Phil1970
– Phil1970, Commented Feb 9, 2019 at 14:01
std::size(x) will only works when std::end(x) - std::begin(x) works so using C++ 17 would not help. C style arrays do not carry their size across function as the decay to pointers. — Phil1970
– Phil1970, Commented Feb 9, 2019 at 14:06

Fureeish · Accepted Answer · 2019-02-09 13:26:42Z

2

Implement std::size yourself:

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

usage:

int main() {
    int arr[] = {1, 2, 3};

    std::cout << "size: " << size(arr);
}

Note that you need to pass a reference to the array, since simply passing T[] will actually make you pass a T* to the first element of the array. That will not preserve any infromation regarding the array's size.

edited Feb 9, 2019 at 13:26

answered Feb 9, 2019 at 13:23

Fureeish

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3 Comments

Eljay Over a year ago

Note that it doesn't chain. You can't have main call void DoSomething(int arr[]) { int sz = size(arr); ... }.

Fureeish Over a year ago

Correct, but the OP seemed like he wanted to implement getArraySize and mentioned that he wants std::size. My note at the bottom should somehow indicate that it won't work if you pass T[], since passing T[] is pretty much the same as passing T*. I also emphasised the usage of a reference to an array to make that point clear.

Eljay Over a year ago

Your post is great, just that the follow up question will be why it doesn't work in the case I mentioned. Then the response will be "You should be using std::vector" (which is what should be used in almost every case in C++ instead of a C array). Which I know you know. :-)

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