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Finding array length -- C++

Started by skulldrudgery Sep 16, 2005 at 9:02 PM 24 replies 149.3k views

Original Post

Sep 16, 2005 at 9:02 PM

Do I HAVE to use a vector? I prefer to use them only when it is necessary. Is there a way to find the length of an array that is passed to a function? I thought I could use something like

while(array) i++;

but that is obviously wrong. (It worked with strings so I figured "oh, what the hell?"). When I create an array inside the same function I can find the size using sizeof(array) / sizeof(char) but I already know the size in that situation.

SiCrane

Sep 16, 2005 at 9:04 PM

No you can't know. Pass the length along with the array or use a vector.

skulldrudgery

Sep 16, 2005 at 9:06 PM

Okie doke. Thanks.

EDIT: How do they find the length with a vector? Just wondering.

Fruny

Sep 16, 2005 at 9:17 PM

Quote:
Original post by skulldrudgery
EDIT: How do they find the length with a vector? Just wondering.

They don't "find" it. They keep track of it. A typical vector implementation would keep a pointer to the beginning of the vector, to the end of the vector (one past the last element) and to the end of the allocated block (one past the last valid address in the block).

The length of the vector would then simply be pEnd - pBegin, and its capacity pEndBlock - pBegin.

skulldrudgery

Sep 16, 2005 at 9:31 PM

So...In other words, shut my uppity n00b mouth and just use the stinking vector.[grin]

Thanks for the info.

Satan666

Sep 16, 2005 at 10:17 PM

#include <vector>int main(){vector<char>Size;Size.push_back('a'); //Size[0] = 'a'Size.push_back('b'); //Size[1] = 'b'Size.push_back('c'); //Size[2] = 'c'int SizesSize = Size.size();//3}

Roboguy

Sep 16, 2005 at 11:20 PM

Quote:
Original post by Anonymous Poster
Quote:

while(array) i++;

Step back for a second. Why does this work for strings and not for other arrays? (Hint: It doesn't work for char arrays. What's the difference between a char array and a string?) Once you figure that out, it may be possible to adapt that solution to your needs.

It should work on a null-terminated char array.

Fruny

Sep 16, 2005 at 11:28 PM

Quote:
Original post by Anonymous Poster
Note that this only works for char arrays and, for a string, will give a different answer from strlen(array). A more robust version is (sizeof array / sizeof *array).

I must voice my disagreement here. (sizeof array / sizeof *array), just like (sizeof array / sizeof char) both suffer from the fatal flaw that they will silently return an erroneous result if array isn't an array name, but a pointer. And given how 'easy' it is for an array to decay to a pointer...

In C++, a safe solution is the following template function:

template size_t N dimension_of(T (&)[N]) { return N; }

or, if you need a version that provides a result at compile-time:

template<size_t N> struct dimension_of_hs { char dummy[N]; };template<class T, size_t N> dimension_of_hs<N> dimension_of_hfn(T (&)[N]);#define dimension_of(X) (sizeof dimension_of_hfn(X))

Both versions will cause a compiler error if you try to pass them pointers.

Usage: dimension_of(array)
Result: The number of elements of the array.

Polymorphic OOP

Sep 17, 2005 at 1:40 AM

Quote:
Original post by Fruny
template<size_t N> struct dimension_of_hs { char dummy[N]; };template<class T, size_t N> dimension_of_hs<N> dimension_of_hfn(T (&)[N]);#define dimension_of(X) (sizeof dimension_of_hfn(X))
Both versions will cause a compiler error if you try to pass them pointers.

Usage: dimension_of(array)
Result: The number of elements of the array.

Actually, technically that wouldn't necessarily work because of padding (the struct size isn't guaranteed to be the size of the array). Instead of returning the array encapsulated in a struct, have the return-type be a reference to an array of char:

template<typename T, ::std::size_t N> char (&dimension_of_hfn(T (&)[N]))[N];

Fruny

Sep 17, 2005 at 2:38 AM

Ooops, my bad.

Mxz

Sep 17, 2005 at 3:13 AM

Heres a slight variation which avoids the need for the struct, and keeps the safety of not compiling if X is only a pointer.

template<class T, size_t N> T decay_array_to_subtype(T (&a)[N]);#define dimension_of(X) (sizeof(X)/sizeof(decay_array_to_subtype(X)))

silvermace

Sep 17, 2005 at 4:17 AM

its true; you really do learn something everyday on these forums.

Fruny

Sep 17, 2005 at 12:34 PM

Sure, it's a bit more complicated, and vectors are preferable in many cases. But it is the safe solution to getting an array's length. And no, it's not possible to get around using a macro, since it's not possible to define new operators in C++. Given that the expression has properly been parenthesed and that the input parameter only appears once in the expression, that code (at least my version, though the sizeof should absorb any function call and thus eliminate side effects in both versions) is safe.

Cacks

Sep 17, 2005 at 12:38 PM

Here is a simple example:

//CODE
int arr[10];
cout << sizeof(arr)/sizeof(int);

sizeof(arr) - returns the size in bytes off the entire array.
sizeof(int) - returns the number of bytes required for one element of the array.
Thus dividing gives the number of elements.

Hope that helps you!

Cocalus

Sep 17, 2005 at 12:58 PM

Quote:
Original post by Cacks
Here is a simple example:

//CODE
int arr[10];
cout << sizeof(arr)/sizeof(int);

sizeof(arr) - returns the size in bytes off the entire array.
sizeof(int) - returns the number of bytes required for one element of the array.
Thus dividing gives the number of elements.

Hope that helps you!

That can break really easily.

#include <iostream>using namespace std;int size(int arr[]){ //I don't work, I always return 1   return sizeof(arr)/sizeof(*arr);}  int main(){    int arraytest[10];    cout << "This works: " << sizeof(arraytest)/sizeof(*arraytest) << endl;    cout << "This doesn't: " << size(arraytest) << endl;    int *dynamicarray = new int[20];    cout << "This doesn't either: " << sizeof(dynamicarray)/sizeof(*dynamicarray) << endl;    delete dynamicarray;    cin.get();}

Cacks

Sep 18, 2005 at 3:35 AM

Yeah, I see now. I forgot that even when declaring a parameter in the form 'arr[]' - 'arr' is a pointer. So using sizeof(arr)/sizeof(type) is basically useless if it only works for statically allocated arrays. Could some1 explain:

//CODE
template T decay_array_to_subtype(T (&a)[N]);
#define dimension_of(X) (sizeof(X)/sizeof(decay_array_to_subtype(X)))

How can this give a compile error if X is only a pointer?

Maybe wrapping an array in a new class with a getLength() function could be a good solution?

Mxz

Sep 18, 2005 at 4:04 AM

Quote:
Original post by Cacks
Yeah, I see now. I forgot that even when declaring a parameter in the form 'arr[]' - 'arr' is a pointer. So using sizeof(arr)/sizeof(type) is basically useless if it only works for statically allocated arrays. Could some1 explain:

//CODE
template T decay_array_to_subtype(T (&a)[N]);
#define dimension_of(X) (sizeof(X)/sizeof(decay_array_to_subtype(X)))

How can this give a compile error if X is only a pointer?

Maybe wrapping an array in a new class with a getLength() function could be a good solution?

It gives a compiler error because an array and a pointer are two distinct types.

A pointer is a variable which contains an address of an element of type T (which may or may not exist).

An array is a contiguous block of N elements of type T.

A variable declared to be a T&, that is “reference to type T”, can only be initialized by an object of type T or by an object that can be converted into a T.

There are no available conversions from type T* (pointer to type T) to type T [N] (array of N elements of type T), so the decay_array_to_subtype function cannot compile when called with a pointer.

JohnBolton

Sep 18, 2005 at 12:34 PM

Quote:
Original post by Anonymous Poster
Anyone got one without the "evil" #define? [looksaround]

This works for me.

    template < typename T, size_t N >    size_t elementsof( T const (&a)[N] )    {        return N;    }

Fruny

Sep 18, 2005 at 12:41 PM

Quote:
Original post by JohnBolton
This works for me.
    template < typename T, size_t N >    size_t elementsof( T const (&a)[N] )    {        return N;    } 

int a[10]; int b[elementsof(a)]; will not compile, nor can us use elementsof(a) as a template argument.

-=HM=-

Feb 23, 2012 at 12:33 AM

_countof(x) implementation in Visual Studio is very similar to what's been posted on this thread already; from memory it looks like this:

template <typename T, std::size_t N> char (* helper(T(&)[N]))[N];
 #define _countof(x) (sizeof (*helper(x)) + 0)

I remember it well because I could never figure out what is that +0 supposed to be doing.

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