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### For loop using the modulo

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:

 ``12345678`` `````` 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.

Just compare the two indices, and set the smaller one to ind1
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
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
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This is my code:
 ``12345678910111213`` ``````#include 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<

Edit: i do not know if the modulo can be done in the for loop step declaration
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This is what i've done. Thanks for your help...... this seems to work for both instances.

 ``123456789101112`` ``````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]; } ``````
 ``12345678910`` ``````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
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