Element count of an array in C++

Question

Let\'s say I have an array arr. When would the following not give the number of elements of the array: sizeof(arr) / sizeof(arr[0])?

I can

Answer 1

_countof(my_array) in MSVC

I can thing of only one case: the array contains elements that are of different derived types of the type of the array.

Elements of an array in C++ are objects, not pointers, so you cannot have derived type object as an element.

And like mentioned above, sizeof(my_array) (like _countof() as well) will work just in the scope of array definition.

Answer 2

Let's say I have an array arr. When would the following not give the number of elements of the array: sizeof(arr) / sizeof(arr[0])?

One thing I've often seen new programmers doing this:

void f(Sample *arr)
{
   int count = sizeof(arr)/sizeof(arr[0]); //what would be count? 10?
}

Sample arr[10];
f(arr);

So new programmers think the value of count will be 10. But that's wrong.

Even this is wrong:

void g(Sample arr[]) //even more deceptive form!
{
   int count = sizeof(arr)/sizeof(arr[0]); //count would not be 10  
}

It's all because once you pass an array to any of these functions, it becomes pointer type, and so sizeof(arr) would give the size of pointer, not array!

---

EDIT:

The following is an elegant way you can pass an array to a function, without letting it to decay into pointer type:

template<size_t N>
void h(Sample (&arr)[N])
{
    size_t count = N; //N is 10, so would be count!
    //you can even do this now:
    //size_t count = sizeof(arr)/sizeof(arr[0]);  it'll return 10!
}
Sample arr[10];
h(arr); //pass : same as before!

Answer 3

Arrays in C++ are very different from those in Java in that they are completely unmanaged. The compiler or run-time have no idea whatsoever what size the array is.

The information is only known at compile-time if the size is defined in the declaration:

char array[256];

In this case, sizeof(array) gives you the proper size.

If you use a pointer as an array however, the "array" will just be a pointer, and sizeof will not give you any information about the actual size of the array.

STL offers a lot of templates that allow you to have arrays, some of them with size information, some of them with variable sizes, and most of them with good accessors and bounds checking.

Answer 4

There are no cases where, given an array arr, that the value of sizeof(arr) / sizeof(arr[0]) is not the count of elements, by the definition of array and sizeof.

In fact, it's even directly mentioned (§5.3.3/2):

.... When applied to an array, the result is the total number of bytes in the array. This implies that the size of an array of n elements is n times the size of an element.

Emphasis mine. Divide by the size of an element, sizeof(arr[0]), to obtain n.

Answer 5

Let's say I have an array arr. When would the following not give the number of elements of the array: sizeof(arr) / sizeof(arr[0])?

In contexts where arr is not actually the array (but instead a pointer to the initial element). Other answers explain how this happens.

I can thing of only one case: the array contains elements that are of different derived types of the type of the array.

This cannot happen (for, fundamentally, the same reason that Java arrays don't play nicely with generics). The array is statically typed; it reserves "slots" of memory that are sized for a specific type (the base type).

Sorry for the trivial question, I am a Java dev and I am rather new to C++.

C++ arrays are not first-class objects. You can use boost::array to make them behave more like Java arrays, but keep in mind that you will still have value semantics rather than reference semantics, just like with everything else. (In particular, this means that you cannot really declare a variable of type analogous to Foo[] in Java, nor replace an array with another one of a different size; the array size is a part of the type.) Use .size() with this class where you would use .length in Java. (It also supplies iterators that provide the usual interface for C++ iterators.)