In the C programming language, you can't determine the array size from the parameter, so you have to pass it in as an extra parameter. The solution could be:
#include <stdio.h>
void swaparrayends(int arr[], int nrElements)
{
int temp = arr[0];
arr[0] = arr[nrElements - 1];
arr[nrElements - 1] = temp;
}
void main()
{
int i;
int myArray[] = { 1,2,3,4,5 };
int nrElements = sizeof(myArray) / sizeof(myArray[0]);
swaparrayends(myArray, nrElements);
for (i = 0; i < nrElements; i++)
{
printf("%d ", myArray[i]);
}
getchar();
}
In higher languages like C# it becomes much simpler:
static void Main(string[] args)
{
int[] myArray = {1, 2, 3, 4, 5};
swaparrayends(myArray);
foreach (var el in myArray)
{
Console.Write(el + " ");
}
Console.ReadLine();
}
static void swaparrayends(int[] arr)
{
int temp = arr[0];
arr[0] = arr.Last();
arr[arr.Length - 1] = temp;
}
22 bits
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Answer:
public class SimpleSquare{
public int num;
private int square;
public SimpleSquare(int number){
num = number;
square = number * number;
}
public int getSquare(){
return square;
}
}
Explanation:
*The code is in Java.
Create a class called SimpleSquare
Declare two fields, num and square
Create a constructor that takes an integer number as a parameter, sets the num and sets the square as number * number.
Since the square is a private field, I also added the getSquare() method which returns the value of the square.
Answer:
is there a pic or something, sorry i cant help at the time being
Explanation:
Answer:
I am a software engineering student
computers have createed an impact in my life because through computers i work, I do research and many more
