C++ negative array indices - c++

Question

I want to loop an array then during each loop I want to loop backwards over the previous 5 elements.
So given this array
int arr[24]={3, 1, 4, 1, 7, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4}
and this nested loop
for(int i=0;i<arr.size;i++)
{
for(int h=i-5; h<i; h++)
{
//things happen
}
}
So, if i=0, second loop would loop last few elements 4,6,2,6,5.
How could you handle this?

Answer 1

I'm assuming that:
You only want to go over previous values (i.e. no wrap around) You
You don't actually want arr to be a multi-dimensional array as suggested
by your choice of tags
You want to include the current i in your five values
This is just a small modification to your code that will do (what I think) you are asking:
#include <math>
int main()
{
int arr[24]={3, 1, 4, 1, 7, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4}
for(int i=0;i<arr.size;i++)
{
for(int h = max(i-4, 0); h < i+1; h++)
{
//things happen
}
}
}
note the h = max(i-4, 0) and h < i+1This will reduce the number of iterations of the inner loop so that it starts from index 0 and loops up through the five values up to and including i. (four values and i). h will always be within bounds.
The case where i==arr.size won't be a problem in the inner loop as the outer loop will terminate before that happens (i is always within bounds).
Edit: I saw this comment:
I want the first element to consider the last final 5 elements of the array though.
in which case, your loops should look like:
for(int i=0;i<arr.size;i++)
{
for(int h=0; h<5; h++)
{
int index = (i + arr.size - h) % arr.size;
//things happen
//access array with arr[index];
}
}
This should do what you want:
When i=0, h=0 index=(0+24-0)%24 which is 0. For h=1 we go one less, index=(0+24-1)%24 = 23 and so on for the next values of h.
The code gets the last 5 values, wrapping round, inclusive of the current value. (so will get 20,21,22,23,0 when i=0, 21,22,23,0,1 when i=1)
If you want the five before, non-inclusive, then inner loop should be:
for(int h=1; h<=5; h++)
here is the current output of the loop as it stands:
i 0 0 0 0 0 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 ... 22 22 22 22 22 23 23 23 23 23
h 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 ... 0 1 2 3 4 0 1 2 3 4
index 0 23 22 21 20 1 0 23 22 21 2 1 0 23 22 3 2 1 0 23 ... 22 21 20 19 18 23 22 21 20 19

Answer 2

I assume you want it to loop around (don't know why). if so, use modulo:
int index = (h + arr.size) % arr.size;

Answer 3

Using the modulo operator.
for (int i = 0; i < arr.size; i++)
{
for (int h = 5; h > 0; h--)
{
const int array_length = sizeof(arr) / sizeof(arr[0]);
int index = (i - h + array_length) % array_length; // Use 'sizeof(arr) / sizeof(arr[0])' to get the size of the array
//things happen
}
}

Answer 4

Is using if statement not an option?
const int array_size = 24;
int arr[array_size] = { 1,3,4,5,...,2 }
for(int i=0;i<array_size;i++)
{
for(int h=i-5; h<i; h++)
{
int arr_index = (h >= 0) ? h : (array_size + h);
//do your things with arr[arr_index]
}
}

Answer 5

you may also start the nested loop with something like:
for(int h=i-min(i,5);h<i;++h)
{
}
which let you process first 5 cells as well. also, if you are dealing with some kind of signal or image processing consider extending arr to have 29 elements with preceding 5 zeros or whatever value would be suitable, and start the first for-loop with 5th element.

Answer 6

Just make an if statement in nested loop. Something like this
for( int h = i-5; h < i; h++ )
{
// do stuff
if( i == 0 )
break;
}