Finding even numbers by using pointers in c++ -

i doing assignment pointers. in 1 of question, asks me find numbers in array , print of them. have use signature given assignment , can not use use & operator or [] notation in function.

signature: void print_evens(int *nums,int length) ;

i know have use,

if(i%2==0)   cout << << endl; 

to find numbers don't know how pointers.

how can pass array main function print_evens since there no parameters array?

thank help.

how can pass array main function print_evens since there no parameters array?

you can never pass array function, regardless of signature. arrays decay pointers first element in such situation.

in case, need dereference pointer value @ current location.

for (int = 0; < length; ++i) {     int current = *(nums + i);     if ((current % 2) == 0)         cout << current << endl; } 

and can pass in array (which decays pointer of course)

#define len 10  int main(void) {     int arr[len];     /* initialize elements of arr */         print_evens(arr, len); }