C# Simple question about arrays

Question

C# code:

int[ ] a = new int[10];
for (int k = 1; k <= 10; k++) {
a[k] = k*2;
}
for (int k = 0; k < 10; k++) {
Console.WriteLine(a[k]);
}
Why do I still get 0 as the output of the first value? From the first for loop I declared k's value as 1 and should be rendering 2...20 instead it shows 0...18

It gives an error of "out of bounds". if I change the first loops condition to <= 9. it still gives me 0 as the initial value and stops at 18

update: so it seems 0 could be a default value given to the "0" position. I would imagine that it would be the last/10th position that would be set with the value of 0. But theres still the question, why doesnt k <= 10 work?

Answer 1

When you declare

int[] a = new int[10];

You declare an integer array of size 10. The valid indexes of this array would be

0 1 2 3 4 5 6 7 8 9

i.e. 10 indexes which starts from 0 and ends with 9.

Go figure.

Answer 2

make last line, console.writeline(a[k+1])<<easy way, not only way.
and 0-10 wont fit in your array only 0-9 so new arr[11] if you want it to hold 0-10.

Answer 3

@emmey I made k = 1 so it can only declare 1-10: (1*2 = 2 = k),(2*2 = 4 = k), etc., instead of 0-10, ie excluding 0 * 2. what im trying to do is output all even natural numbers ie 2-20. I think the last line works fine. the first for loop is the one giving an error

thanks for answering

Answer 4

int[ ] a = new int[11];
for (int k = 1; k <= 10; k++) {
a[k] = k*2;
Console.WriteLine(a[k]);

works for that without the second loop.

Answer 5

@Hatsy Rei yeah I get it now. I thought a[k] = k*2; was "enumerating" the values in the array, like: new int[10] {k};
I actually needed to add another operation to get the desired result: a[k] = (k+1)*2;

so in actuality it means:
(k+1)*2 = value is set/index it to a[0,1,2...9] "position"

sololearn never really explained it. maybe, yet.

Answer 6

c# code

int[ ] arr = new int[7];
for(int k=0;k<7 ;k++)
{
Console.WriteLine(array[k]);
}