C#: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

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.

.

int acc = 0; foreach (int i in new int[] {2,7,1,1}) { acc += i; } Console.WriteLine(acc);

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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.

.

int acc = 0; foreach (int i in new int[] {2,7,1,1}) { acc ++; } Console.WriteLine(acc);

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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.

.

int acc = 0; foreach (int i in new int[] {2,7,1,1}) { if (i==1) acc ++; } Console.WriteLine(acc);

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Using an accumulator to multiply

Print the product of the numbers.

• We initialise to 1. (If you initialise to 0 then the result will always be 0.)
• We multiply the accumulator.
• The value printed is the product: 1*2*7*1*1

.

int acc = 1; for (int i : new int[]{2,7,1,1}) { acc *= i; } System.out.println(acc);

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Using an accumulator to find the maximum

Print the largest number in the list.

• We initialise to 0.
• We take the max.
• The value printed is the largest:

At each stage the accumulator stays the same or gets bigger (set to the current value).

max(max(max(max(0,2),7),1),1)

What about smaller numbers?

If all of the numbers in the list were negative, for example {-2,-7,-1,-1} then this program would print 0 instead of the correct answer -1. You can fix that by starting the accumulator at the smallest possible integer: int acc = Integer.MIN_VALUE; for (int i : new int[]{2,7,1,1}) { acc = Math.max(acc,i); } System.out.println(acc);

.

int acc = 0; foreach (int i in new int[] {2,7,1,1}) { acc = Math.Max(acc,i); } Console.WriteLine(acc);

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Using an accumulator to concatenate

Concatenate all items in the list

Notice that we are using a list of strings this time.

• We initialise to the empty string "".
• We concatenate the next value (stick at the end).

WARNING - Using an accumulator like this can use a great deal of memory. You will find using a C#:StringBuilder is more efficient.

System.Text.StringBuilder acc = new System.Text.StringBuilder();
foreach (String i in new String[] {"two","seven","one","one"})
{
  acc.Append(i);
}
Console.WriteLine(acc);

Also you can use the method String.Join for this operation.

.

String acc = ""; foreach (String i in new String[] {"two","seven","one","one"}) { acc += i; } Console.WriteLine(acc);

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Using two accumulators to find the mean

Print the mean of all items in the list.

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

.

int count = 0; int sum = 0; for (int i : new int[]{3,7,1,1}) { count++; sum += i; } System.out.println((float)sum/count);

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