Passing array as function parameter in C++

Question

I am aware that an array can be passed to a function in quite a few ways.

#include <iostream>
#include <utility>

using namespace std;

pair<int, int> problem1(int a[]);

int main()
{
    int a[] = { 10, 7, 3, 5, 8, 2, 9 };
    pair<int, int> p = problem1(a); 
    cout << "Max =" << p.first << endl;
    cout << "Min =" << p.second << endl;
    getchar();
    return 0;
}

pair<int,int> problem1(int a[])
{
    int max = a[0], min = a[0], n = sizeof(a) / sizeof(int);

    for (int i = 1; i < n; i++)
    {
        if (a[i]>max)
        {
            max = a[i];
        }
        if (a[i] < min)
        {
            min = a[i];
        }

    }

    return make_pair(max,min);


}

My code above passes only the first element while it should be passing an array (or technically, a pointer to the array) and hence, the output is 10, 10 for both max and min (i.e. a[0] only).

What am I doing wrong, I guess this is the correct way.

`n = sizeof(a) / sizeof(int)` doesn't do what you think it does. You need to pass in N or use a non-type template argument. — WhozCraig, Jul 10 '15 at 17:36
Btw your compiler would have told you the problem if you had enabled a reasonable warning level: http://coliru.stacked-crooked.com/a/309fb24cde22a907 — Baum mit Augen, Jul 10 '15 at 17:42
I realize the problem is probably academic, but you could just as well use [`std::minmax_element`](http://en.cppreference.com/w/cpp/algorithm/minmax_element) for this and belay the function entirely. [See it live](http://ideone.com/PtBaMf). — WhozCraig, Jul 10 '15 at 18:02
@adrian008 _"... but a flurry of downvotes is undeserved"_ why do you think so? That exactly was meant as written. — πάντα ῥεῖ, Jul 10 '15 at 18:52

NathanOliver · Accepted Answer · 2015-07-10T18:16:30.967

5

The contents of the array are being passed to the function. The problem is:

n = sizeof(a) / sizeof(int)

Does not give you the size of the array. Once you pass an array to a function you can't get its size again.

Since you aren't using a dynamic array you can use a std::array which does remember its size.

You could also use:

template <int N>
void problem1(int (&a) [N]) 
{
    int size = N;
    //...
}

edited Jul 10 '15 at 18:16

answered Jul 10 '15 at 17:37

NathanOliver

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No, you're not passing the array. You're passing a pointer. See [my answer](http://stackoverflow.com/a/31347098/827263). – Keith Thompson Jul 10 '15 at 18:13

score 4 · Answer 2 · answered Jul 10 '15 at 17:44

No, you simply cannot pass an array as a parameter in C or C++, at least not directly.

In this declaration:

pair<int, int> problem1(int a[]);

even though a appears to be defined as an array, the declaration is "adjusted" to a pointer to the element type, so the above really means:

pair<int, int> problem1(int* a);

Also, an expression of array type is, in most contexts, implicitly converted to a pointer to the array's initial element. (Exceptions include an array as the operand of sizeof or unary &). So in a call to the above function:

int arr[10];
problem1(arr);

the array expression arr is equivalent to &arr[0], and that address (pointer value) is what's passed to the function.

Of course you can write code that does the equivalent of passing an array. You can make the array a member of a structure (but then it has to be of fixed length). Or you can pass a pointer to the initial element and pass a separate parameter containing the actual length of the array object.

Or you can use one of the C++ standard library classes that implement array-like data structures; then the length can be taken directly from the parameter.

I highly recommend reading section 6 of the comp.lang.c FAQ, which covers arrays and pointers. It's applicable to C++ as well (though it doesn't mention the C++ standard library).

score 0 · Answer 3 · answered Jul 10 '15 at 17:42

In C++ language a function parameter declared as int a[] is immediately interpreted as and is equivalent to int *a parameter. Which means that you are not passing an array to your function. You are passing a pointer to the first element of an array.

Trying to apply the sizeof(a) / sizeof(int) technique to a pointer is useless. It cannot possibly produce the size of the argument arraay.

score 0 · Answer 4 · answered Jul 10 '15 at 17:56

One alternative that hasn't been mentioned is writing your code as a template, and passing the array by reference so the template can deduce the size of the array:

template <class T, size_t n>
pair<T, T> problem1(T(&a)[n]) {
    T max = a[0], min = a[0];

    for (size_t i = 1; i < n; i++) {
        if (a[i]>max) {
            max = a[i];
        }
        if (a[i] < min) {
            min = a[i];
        }
    }
    return make_pair(max, min);
}

Note, however, that this will only work if you pass a real array, not a pointer. For example, code like this:

int *b = new int[10];

for (int i = 0; i < 10; i++)
    b[i] = rand();

auto result = problem1(b);

...won't compile at all (because we've defined problem1 to receive a reference to an array, and b is a pointer, not an array).

Passing array as function parameter in C++

4 Answers4

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