Why does the length function return 2?

Question

I have written a function size which takes integer array as an argument.

int length(int a[])
{

    return  sizeof(a)/sizeof(int);

}

int main()
{
    int a[] = {1,3,5,6,9,4,2,1,0,0};

    int len = sizeof(a)/sizeof(int);

    cout << len;  // This correctly prints 10 .

    len =  size(a);

    cout << len;  // But this print 2 .why ??

    return 0;
}

Can someone explain me this behaviour? Thanks.

`int length(int a[])` is the same as `int length(int* a)`. Don't use that notation since, as you can see, it is misleading. — R. Martinho Fernandes, Jul 04 '13 at 13:12
This *will* explain it: http://stackoverflow.com/questions/4810664/how-do-i-use-arrays-in-c — chris, Jul 04 '13 at 13:13
@GrijeshChauhan not necessarily. Could very well be a 64bit system with 64bit pointer, 32bit int. Or maybe 60bit pointer, 30bit int, 10bit char. You can just tell the ratio pointer/int, nothing else — Arne Mertz, Jul 04 '13 at 13:16
@R.MartinhoFernandes Can you explain me why it is misleading ? I am not able to understand. — Thinker, Jul 04 '13 at 13:17
Inside main, compiler knows what `a` is. In the `length` function, it doesnt, because it is just a pointer parameter. — ondrejdee, Jul 04 '13 at 13:18
@R.MartinhoFernandes Why 64-bit? That's likely but not definite — SomeWittyUsername, Jul 04 '13 at 13:18
Then what else do you need me to explain to show it is misleading? It (wrongly) lead you to think you are passing an array, and then you drew the wrong conclusions because you are actually passing a pointer. — R. Martinho Fernandes, Jul 04 '13 at 13:20
@Thinker, there's a [good explanation](http://stackoverflow.com/a/1328246/2417578) of what's going on in the first of the links that this question duplicates. Arrays as function arguments work differently from your expectations. — sh1, Jul 04 '13 at 13:24

score 4 · Accepted Answer · answered Jul 04 '13 at 13:14

4

The reason you get 2 is because sizeof(int *) is twice as large as sizeof(int), and arrays decay into pointers when passed into a function.

There are various ways to work around this. In C++, you could use a std::vector<int> a = { ... }; instead, which would solve the problem (by calling the a.size() to get the size, as sizeof(a) wouldn't work).

answered Jul 04 '13 at 13:14

Mats Petersson

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Or pass the array as `int (&)[N]` – David G Jul 04 '13 at 13:17
@Mats can you give me some link regarding std:vector tutorial. I have no idea about that – Thinker Jul 04 '13 at 13:18
http://www.cplusplus.com/reference/vector/vector/ Not quite a tutorial, but perhaps a starting point. I'm sure google with "tutorial C++ vector" may give you some other links that at least some may be good - I haven't been "learning" about C++ (at that level at least), so not sure where a good tutorial site is. – Mats Petersson Jul 04 '13 at 13:22
@Thinker you should use a good beginners' C++ textbook. Things like `std::vector` are the very basic building blocks of the standard library as part of the language. It's a bread-and-butter `feature` like `double`, for-loops and the like. – Arne Mertz Jul 04 '13 at 13:42

score 0 · Answer 2 · answered Jul 04 '13 at 13:16

Because sizeof(a) is returning the size of the pointer which will be either 4 bytes or 8 bytes depending upon whether your app is 32-bit or 64-bit. So you would need to maintain the length of your array using a separate variable.

Or, better still, I would recommend using one of the standard collection classes, like std::vector for example since this will maintain the length automatically.

Why does the length function return 2?

2 Answers2