The second to last is
array[array.Length - 2];
not
array[array.Length - 1];
... since arrays are zero-based. However, you should check if it contains enough elements.
题
Hi I wrote a method/array I am trying to make my code swap the second number and second to last number in the index but the only number that's is swapping was my second number and my last, if you could help a newbie I would be very thank full.
static void Main(string[] args)
{
Console.WriteLine("\n=====================\n");
DisplayArray(A);
Console.WriteLine("\nSwapping Second and Second to last Number");
SwapSecAndSecLast(A);
DisplayArray(A);
Console.ReadLine();
}
static void SwapSecAndSecLast(int[] array)
{
int temp = array[1];
array[1] = array[array.Length - 1];
array[array.Length - 1] = temp;
}
static void DisplayArray(int[] array)
{
Console.WriteLine("\n=====================\n");
for (int i = 0; i < array.Length; i++)
{
Console.Write("{0} ",array[i]);
}
Console.WriteLine("\n=====================\n");
}
解决方案
The second to last is
array[array.Length - 2];
not
array[array.Length - 1];
... since arrays are zero-based. However, you should check if it contains enough elements.
其他提示
Arrays are 0 based, so array[array.Length - 1] is the last element in the array. If you want the second to last, use array[array.Length - 2]