# Program to print the elements of an array present on even position

Program to print the elements of an array present on even position on fibonacci, factorial, prime, armstrong, swap, reverse, search, sort, stack, queue, array, linkedlist, tree, graph etc.

## Q. Program to print the elements of an array present on even position.

### Explanation

In this program, we need to print the element which is present in even position.

Even positioned element can be found by traversing the array and incrementing the value of i by 2.

In the above array, elements on even position are b and d.

### Algorithm

1. Declare and initialize an array.
2. Calculate the length of the declared array.
3. Loop through the array by initializing the value of variable "i" to 1 (because first even positioned element lies on i = 1) then incrementing its value by 2, i.e., i=i+2.
4. Print the elements present in even positions.

### Python

```#Initialize array
arr = [1, 2, 3, 4, 5];

print("Elements of given array present on even position: ");
#Loop through the array by incrementing the value of i by 2

#Here, i will start from 1 as first even positioned element is present at position 1.
for i in range(1, len(arr), 2):
print(arr[i]);
```

Output:

```Elements of given array present on even position:
2
4
```

### C

```#include

int main()
{
//Initialize array
int arr[] = {1, 2, 3, 4, 5};
//Calculate length of array arr
int length = sizeof(arr)/sizeof(arr[0]);
printf("Elements of given array present on even position: \n");
//Loop through the array by incrementing value of i by 2
//Here, i will start from 1 as first even positioned element is present at position 1.
for (int i = 1; i < length; i = i+2) {
printf("%d\n", arr[i]);
}
return 0;
}
```

Output:

```Elements of given array present on even position:
2
4
```

### JAVA

```public class EvenPosition {
public static void main(String[] args) {

//Initialize array
int [] arr = new int [] {1, 2, 3, 4, 5};

System.out.println("Elements of given array present on even position: ");
//Loop through the array by incrementing value of i by 2
//Here, i will start from 1 as first even positioned element is present at position 1.
for (int i = 1; i < arr.length; i = i+2) {
System.out.println(arr[i]);
}
}
}
```

Output:

```Elements of given array present on even position:
2
4
```

### C#

```using System;
public class EvenPosition
{
public static void Main()
{
//Initialize array
int [] arr = new int [] {1, 2, 3, 4, 5};

Console.WriteLine("Elements of given array present on even position: ");
//Loop through the array by incrementing value of i by 2
//Here, i will start from 1 as first even positioned element is present at position 1.
for (int i = 1; i < arr.Length; i = i+2) {
Console.WriteLine(arr[i]);
}
}
}
```

Output:

```Elements of given array present on even position:
2
4
```

### PHP

```<!DOCTYPE html>
<html>
<body>
<?php
//Initialize array
\$arr = array(1, 2, 3, 4, 5);

print("Elements of given array present on even position: <br>");
//Loop through the array by incrementing value of i by 2
//Here, i will start from 1 as first even positioned element is present at position 1.
for (\$i = 1; \$i < count(\$arr); \$i = \$i+2) {
print(\$arr[\$i] . "<br>");
}
?>
</body>
</html>
```

Output:

```Elements of given array present on even position:
2
4
```

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