Passing array as function parameter in C++ -

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• when function has specific-size array parameter, why replaced pointer? 3 answers

i aware array can passed function in quite 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);      (int = 1; < 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 first element while should passing array (or technically, pointer array) , hence, output 10, 10 both max , min (i.e. a[0] only).

what doing wrong, guess correct way.

the contents of array being passed function. problem is:

n = sizeof(a) / sizeof(int) 

does not give size of array. once pass array function can't size again.

since aren't using dynamic array can use std::array remember size.

you use:

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