# Using an Accumulator

When using an accumulating variable there are three stages:

- Initialise
- We declare our accumulator and set it to an initial values.
- Update
- Inside the loop we update the accumulator - we take into account the current data item.
- Output/use
- After the loop our accumulator contains the required value. We use it or output it.

## Contents

- 1 Using an accumulator to add.
- 2 Using an accumulator to count.
- 3 Using an accumulator to count on a condition.
- 4 Using an accumulator to multiply.
- 5 Using an accumulator to find the maximum.
- 6 Using an accumulator to concatenate.
- 7 Using two accumulators to find the mean.
- 8 Using an accumulator to calculate.

## Using an accumulator to add.

Print the total of all of the numbers

- We initialise to 0.
- We add to the accumulator.
- The value printed is the sum 0+2+7+1+1.

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## Using an accumulator to count.

Print the number of items in the list.

- We initialise to 0.
- We increment the accumulator.
- The value printed is the count: 0+1+1+1+1.

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## Using an accumulator to count on a condition.

Count the number of 1s in the list

- We initialise to 0.
- We increment the accumulator only if the current value equals 1.

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## Using an accumulator to multiply.

- We initialise to 1.
- We multiply the accumulator.
- The value printed is the product: 1*2*7*1*1

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## Using an accumulator to find the maximum.

- We initialise to 0.
- We take the max.
- The value printed is the largest:
max(max(max(max(0,2),7),1),1)

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## Using an accumulator to concatenate.

- We initialise to "".
- We concatenate the next value.
- Each time i gets converted to a string and is put at the end.
- The answer is ""+2+7+1+1

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## Using two accumulators to find the mean.

- We initialise both to 0.
- We add to the sum and increment the count.
- We divide the sum by the count to get the average.

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## Using an accumulator to calculate.

- We initialise to 0.
- We multiply by the radix and add the next digit.
- By the end the number 2 has been multiplied by 10 three times, the number has been multiplied by 10 twice...
- 10*(10*(10*2+7)+1)+1

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