For loop using the modulo

[jazpearson]

My brain doesn't seem to be working at all today.

Here's my problem. Imagine having an array of integers, but i want to add up these integers from a pair of indices.

So, let's imagine we have an integer array containing 10 items, and the two indices are given by 3 and 8. To do the sum, we would have:

int a[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int ind1 = 3, ind2 = 8;
int total = 0;

for (int i = ind1; i <= ind2; i++)
  total += a[i];

Of course, this is simple. However, what happens if i want to go from indices of 8 to 3?
That is i want to calculate 9+10+1+2+3+4........

I could write two for loops to handle this situation, but i'm sure there's a way to do this in one loop with the help of the modulo. I just can't remember.

[Script Coder]

Just compare the two indices, and set the smaller one to ind1

[jazpearson]

That won't do what i want though.

That would add up the values as before: 4+5+6+7+8+9
I want: 9+10+1+2+3+4

[Script Coder]

I see, then you would have to inside your for loop check to see if you are one the last number and if so set i=0, or always modulo i by the length of the array

[Script Coder]

This is my code:

#include <iostream>
using namespace std;
int main() {
	int a[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
	int ind1 = 8, ind2 = 3;
	int total = 0;

	for (int i = ind1; i != ind2+1; i++) {
		i%=10;
		total += a[i];
	}
	cout<<total;
}

Edit: i do not know if the modulo can be done in the for loop step declaration

[jazpearson]

This is what i've done. Thanks for your help...... this seems to work for both instances.

int a[10] = {1,2,3,4,5,6,7,8,9,10};
int ind1 = 8, ind2 = 3;
int total = 0;

int num = (ind1 < ind2) ? ind2 - ind1 + 1 : 10 - ind1 + ind2 + 1;

for (int i = 0; i < num; i++)
{
  int index = (i + ind1) % 10;
  total += a[index];
}

[vlad from moscow]

const int N = 10;

for ( int i = 0; i < N; i++ )
{
	for ( int j = i, k = N; k-- != 0; j + 1 == N ? j = 0 : ++j )
	{
		std::cout << j << ' ';
	}
	std::cout << std::endl;
}

Or the inner loop can be written as

for ( int j = i, k = N; k-- != 0; j = ( j + 1 ) % N )

0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 0
2 3 4 5 6 7 8 9 0 1
3 4 5 6 7 8 9 0 1 2
4 5 6 7 8 9 0 1 2 3
5 6 7 8 9 0 1 2 3 4
6 7 8 9 0 1 2 3 4 5
7 8 9 0 1 2 3 4 5 6
8 9 0 1 2 3 4 5 6 7
9 0 1 2 3 4 5 6 7 8