Integer Arithmetic Operations on Large Numbers

Maximum positive value that can be stored by integer type variable in ABAP is 2147483647.

For values beyond maximum limit of integer variable, workarounds can be used.

One such example is discussion MOD operation fail for large numbers started by Starlet Abraham

I was able to calculate correct result of 512^17 MOD 2773 using the methods detailed below.

I have written this program to do following operations on large integers.

• Addition
• Subtraction
• Multiplication
• Division
• Mod
• Power

Full code snippet is posted at the end of this document.

Program has its flaws, but it works.

I would briefly go through how the logic works.

How are big numbers stored?

Local class lcl_bignum has instance attributes that store big numbers.

LV_STRING - Number stored as string type.

LT_INT - an internal table of integers. The string is broken into substrings of size 4 (base 10000), and those substrings ( as int4 ) are stored in internal table.

LV_SIGN - integer type to store the sign of number. +1 for positive, -1 for negative.

For example, number 1234567890 will be stored in internal table as

Why is number broken into base 10000?

Base 10000 is safe considering the capacity of integer variable type in ABAP.

Computing 9999 * 9999 results in 99980001, which is within the limits of integer type.

On the other hand, 99999 * 99999 results in 999980001, which would lead to overflow exception.

How to create instance and store big number?

Below code snippets can be used to create instance of big number.

The constructor parameters are options.

CREATE OBJECT lr3.

--

Above code will create instance with zero value in it.

CREATE OBJECT lr2
  EXPORTING
    i_value = '533335555511111'.

--

Above code will create instance and populate lv_string and lt_int by breaking down input string.

Sign of number is positive by default.

Input string should have only numbers in it.

CREATE OBJECT lr1
  EXPORTING
    i_value = '6584389034291301828999'
    i_negative = 'X'.

--

Using above code, 'X' can be passed i_negative flag if we have a negative number.

How to add numbers?

Instance method Add( ) can be used for adding numbers.

It accepts 2 object reference that need to be added, and result is stored in object whose method gets called.

We are adding the numbers starting lowest weightage (LT_INT[1]), and carrying forward extra in case of overflow.

For example, we add 12345678 and 22223333. Program would add (5678,3333) first, and then (1234,2222).

For A = B + C, below code can be used.

lr3->add( i_obj1 = lr1
          i_obj2 = lr2 ).

--

For A = A + B, below code can be used.

lr3->add( i_obj1 = lr3
          i_obj2 = lr2 ).

--

How to subtract numbers?

Instance method Subtract( ) can be used for subtracting numbers.

The usage is same as Add( ) method.

Looking at code, you would see that Add( ) and Subtract( ) are calling each other.

It is done the tackle how absolute values are treated depending on their sign.

For example, A = B + C, the values of B and C will be added, and common sign will be retained if B and C have same sign. Both positive, or both negative.

If the signs of B and C are different, we essentially have to keep sign of B, and calculate |B| - |C| (absolute values).

How to multiply numbers?

Instance method Multiply( ) can be used.

It accepts 2 object references, and stores result in calling object.

I would explain the logic using example. It is similar to the method used in elementary school.

For elementary example of multiplying 123 and 456, below logic is used.

123

456

===

3*6 + (3*5+2*6) + (3*4+2*5+1*6) + (2*4+1*5) + 1*4

Same thing is done in program at base 10000 level.

At every stage we are carrying forward the overflow value.

How to divide numbers?

For division, I am guessing the answer by multiplying.

Something like binary search, I start with 1 as answer, keep on multiplying it by 10000 still answer crosses dividend.

Once the guess surpasses dividend, I take a step back, and start multiplying by square root of 10000.

A point comes when multiplier is reduced to 1, and multiplication can no longer be used to get closer to answer.

I then use addition to get closer to answer. Instead of square root, divide by 2 to get next "jump value".

A dry run using elementary example would be something like this.

451

4

---

How to do calculate remainder (MOD operation)

Instance method Mod( ) can be used, which internally uses Divide( ), Multiply( ) and Subtract( ).

To calculate A = B MOD C, program does something like:

A = B - C * ( B DIV C )

Here we are reusing the code written in previous methods.

How to calculate Power?

Instance method Power( ) can be used, which internally uses Multiply( ) and Subtract( ).

For calculating A = B POWER C, program starts with 1 as answer, uses loop to multiply B to answer, C number of times.

Full code snippet

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