This algorithm is about to store strings from Array A to Array B by storing "A", "B" to Index 8 and Index 9 I really initiate to make the array size of B to be 10 because later I will put some other things there.
My partial code:
string[] A = new string[]{"A","B"}
string[] B = new string[10];
int count;
for(count = 0; count < A.length; count++)
{
B[count] = A[count]
}
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In the real sense it has no meaning or full form. It was developed by Dennis Ritchie and Ken Thompson at AT&T bell Lab. First, they used to call it as B language then later they made some improvement into it and renamed it as C and its superscript as C++ which was invented by Dr.
So you want to increment every index with 2:
string[] A = new string[] { "A", "B", "C", "D" };
string[] B = new string[A.Length + 2];
for (int i = 0; i < A.Length; i++)
{
B[i + 2] = A[i];
}
Demo
Index: 0 Value:
Index: 1 Value:
Index: 2 Value: A
Index: 3 Value: B
Index: 4 Value: C
Index: 5 Value: D
Edit: So you want to start with index 0 in B and always leave a gap?
string[] A = new string[] { "A", "B", "C", "D" };
string[] B = new string[A.Length * 2 + 2]; // you wanted to add something other as well
for (int i = 0; i/2 < A.Length; i+=2)
{
B[i] = A[i / 2];
}
Demo
Index: 0 Value: A
Index: 1 Value:
Index: 2 Value: B
Index: 3 Value:
Index: 4 Value: C
Index: 5 Value:
Index: 6 Value: D
Index: 7 Value:
Index: 8 Value:
Index: 9 Value:
Update " Is there any alternative coding aside from this?"
You can use Linq, although it would be less readable and efficient than a simple loop:
String[] Bs = Enumerable.Range(0, A.Length * 2 + 2) // since you want two empty places at the end
.Select((s, i) => i % 2 == 0 && i / 2 < A.Length ? A[i / 2] : null)
.ToArray();
Final Update according to your last comment(start with index 1 in B):
for (int i = 1; (i-1) / 2 < A.Length; i += 2)
{
B[i] = A[(i-1) / 2];
}
Demo
Index: 0 Value:
Index: 1 Value: A
Index: 2 Value:
Index: 3 Value: B
Index: 4 Value:
Index: 5 Value: C
Index: 6 Value:
Index: 7 Value: D
Index: 8 Value:
Index: 9 Value
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