通过使用C ++中的指针查找偶数

Question

I am doing an assignment about pointers. In one of the question, it asks me to find even numbers in an array and print all of them. I also have to use the signature that is given by the assignment and I can not use use the & operator or [] notation in my function.

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

I know I have to use,

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

to find the even numbers but I don't know how to do it with pointers.

How can I pass an array from main function to the print_evens since there are no parameters for array?

Thank you for any help.

Answer 1

How can I pass an array from main function to the print_evens since there are no parameters for array?

You can never pass an array to a function, regardless of the signature. Arrays will decay to pointers to their first element in such a situation.

In any case, you need to dereference the pointer to get the value at the current location.

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

And you can pass in your array (which decays to a pointer of course)

#define len 10

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

Answer 2

Consider the following code :

void print_evens(int *nums,int length){

    for(int i = 0 ; i<length ; i++){
        int currNum = *(nums+i);
        if(currNum%2 != 1)
            cout<<currNum<<endl;    
    }
}

it will do what you want.

But how it is done is it because nums is a pointer so you could simply iterate through it by adding number to it. Every time you think you are working with array, in fact it is pointer to the array that you works with.

according to the standard we have :

6.5.2.1 Array subscripting

A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))).

so even in arrays when you say nums[i] it means *(nums+i), and with using this fact you could simply work with pointers.

Answer 3

void print_evens(int *nums,int length)
{
   for (int i = 0; i < length; ++i) {
      if ((nums[i] % 2) == 0)
        cout<< nums[i] << endl;
   }
}

int main(void) {
    int arr[<size of array you want>];
    //Assign values to array    
    print_evens(arr, len);
}

You can use pointers as arrays for your convenience. No one will stop you :-)

Answer 4

Here is a demonstrative program

#include <iostream>

void print_evens( int *nums, int length )
{
    for ( int *p = nums; p != nums + length; ++p )
    {
        if ( *p % 2 == 0 ) std::cout << *p << ' ';
    }
}

int main() 
{
    const int N = 10;
    int nums[N] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };

    print_evens( nums, N );

    std::endl( std::cout );

    return 0;
}

The output is

2 4 6 8 10 

Of course it would bebetter to declare the function like

void print_evens( const int *nums, size_t length );

that constant arrays also could be outputed.