Passing array as function parameter in C++ [duplicate] - c++

Question

This question already has answers here:
When a function has a specific-size array parameter, why is it replaced with a pointer?
(3 answers)
Closed 7 years ago.
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.

Answer 1

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;
//...
}

Answer 2

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).

Answer 3

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.

Answer 4

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).