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Understanding Ada's Remainder Operators: rem vs. mod

Understanding Ada's Remainder Operators: rem vs. mod

Explore the nuances of Ada's rem and mod operators, their mathematical definitions, and practical implications for integer arithmetic, especially with negative numbers.

Ada provides two distinct remainder operators: rem (remainder) and mod (modulo). While they often produce the same result for positive operands, their behavior diverges significantly when negative numbers are involved. Understanding this difference is crucial for writing correct and predictable arithmetic logic in Ada programs, especially in contexts like cyclic buffers, hash functions, or cryptographic algorithms where the sign of the result matters.

The rem Operator: Symmetric Remainder

The rem operator in Ada computes the remainder of a division, where the sign of the result is the same as the sign of the dividend (the first operand). This behavior is consistent with the definition of remainder in many programming languages (e.g., C, C++, Java's % operator). Mathematically, for integers A and B (where B is not zero), A = B * (A / B) + (A rem B), and the absolute value of (A rem B) is less than the absolute value of B. The division A / B truncates towards zero.

declare
   A : Integer;
   B : Integer;
begin
   A := 10;
   B := 3;
   -- 10 = 3 * 3 + 1
   Put_Line ("10 rem 3 = " & Integer'Image (A rem B)); -- Output: 1

   A := -10;
   B := 3;
   -- -10 = 3 * (-3) + (-1)
   Put_Line ("-10 rem 3 = " & Integer'Image (A rem B)); -- Output: -1

   A := 10;
   B := -3;
   -- 10 = (-3) * (-3) + 1
   Put_Line ("10 rem -3 = " & Integer'Image (A rem B)); -- Output: 1

   A := -10;
   B := -3;
   -- -10 = (-3) * 3 + (-1)
   Put_Line ("-10 rem -3 = " & Integer'Image (A rem B)); -- Output: -1
end;

Demonstration of Ada's rem operator with various positive and negative operands. Notice the sign of the result matches the dividend.

The mod Operator: Mathematical Modulo

The mod operator in Ada implements the mathematical definition of the modulo operation, where the result always has the same sign as the divisor (the second operand). This is particularly useful in applications requiring results within a specific range, such as array indexing for circular buffers, where you want a positive index. Mathematically, A = B * N + (A mod B) for some integer N, where 0 <= (A mod B) < B if B > 0, and B < (A mod B) <= 0 if B < 0. The division A / B truncates towards negative infinity.

declare
   A : Integer;
   B : Integer;
begin
   A := 10;
   B := 3;
   -- 10 = 3 * 3 + 1
   Put_Line ("10 mod 3 = " & Integer'Image (A mod B)); -- Output: 1

   A := -10;
   B := 3;
   -- -10 = 3 * (-4) + 2
   Put_Line ("-10 mod 3 = " & Integer'Image (A mod B)); -- Output: 2

   A := 10;
   B := -3;
   -- 10 = (-3) * (-3) + 1
   Put_Line ("10 mod -3 = " & Integer'Image (A mod B)); -- Output: -2
   -- Note: 10 mod -3 = -2 because the result must be between -3 and 0.
   -- 10 = (-3) * (-4) + (-2) is incorrect for mod. It should be
   -- 10 = (-3) * (-3) + 1. However, the definition implies
   -- 10 = (-3) * N + (A mod B) such that B < (A mod B) <= 0
   -- Here, N would be -3, and 10 = -3 * (-3) + 1. This means (A mod B) would be 1.
   -- This is the common interpretation, but Ada's definition is strict:
   -- A = B * N + (A mod B) and B < (A mod B) <= 0 if B < 0.
   -- So, for 10 mod -3, the result must be -2, because 10 = (-3) * (-4) + (-2).

   A := -10;
   B := -3;
   -- -10 = (-3) * 4 + 2
   Put_Line ("-10 mod -3 = " & Integer'Image (A mod B)); -- Output: -1
   -- Similar to above, -10 mod -3 = -1 because -10 = (-3) * 3 + (-1).
end;

Examples of Ada's mod operator. Observe how the result's sign always matches the divisor, and the range constraints.

Visualizing the Difference

To solidify the understanding, let's visualize the results of rem and mod for a dividend of -10 and a divisor of 3. This clearly illustrates how the two operators produce different values and why their distinct definitions are important.

A comparison diagram showing the results of -10 rem 3 and -10 mod 3. For -10 rem 3, an arrow points from -10 to -1 on a number line, indicating the result -1. For -10 mod 3, an arrow points from -10 to 2 on a number line, indicating the result 2. The number line spans from -12 to 3. The rem result has the sign of the dividend, while mod result has the sign of the divisor (positive 3, so positive 2).

Visual comparison of rem and mod for -10 divided by 3.

When to Use Which Operator

The choice between rem and mod depends entirely on the desired behavior and the problem you are trying to solve:

  • Use rem when you need a remainder whose sign matches the dividend. This is often the case in general-purpose arithmetic where the concept of 'remainder' is tied to the sign of the number being divided, or when mimicking the behavior of % in C-like languages.

  • Use mod when you need a result that is strictly non-negative (if the divisor is positive) or strictly non-positive (if the divisor is negative). This is essential for operations like:

    • Circular Buffers/Arrays: Ensuring an index always falls within 0 to N-1 for a buffer of size N.
    • Hash Functions: Producing a hash value within a specific positive range.
    • Time Calculations: For example, calculating the day of the week, where the result must be 0..6.

Failing to choose the correct operator can lead to subtle bugs, especially when inputs can be negative.

Practical Application: Circular Buffer Indexing

Consider a circular buffer where you want to access elements using an index that wraps around. If the index can potentially become negative (e.g., current_index - 1), mod is the correct operator to ensure the index remains valid and positive.

declare
   BUFFER_SIZE : constant Integer := 5;
   Buffer      : array (0 .. BUFFER_SIZE - 1) of Integer;
   Current_Index : Integer := 0;
   New_Index     : Integer;
begin
   -- Initialize buffer (for demonstration)
   for I in Buffer'Range loop
      Buffer (I) := I * 10;
   end loop;

   Put_Line ("Buffer: " & Integer'Image (Buffer(0)) & ", " & Integer'Image (Buffer(1)) & ", " & Integer'Image (Buffer(2)) & ", " & Integer'Image (Buffer(3)) & ", " & Integer'Image (Buffer(4)));

   -- Moving forward
   Current_Index := 4;
   New_Index := (Current_Index + 1) mod BUFFER_SIZE;
   Put_Line ("(4 + 1) mod 5 = " & Integer'Image (New_Index)); -- Expected: 0

   -- Moving backward, potentially resulting in a negative intermediate
   Current_Index := 0;
   New_Index := (Current_Index - 1 + BUFFER_SIZE) mod BUFFER_SIZE;
   Put_Line ("(0 - 1 + 5) mod 5 = " & Integer'Image (New_Index)); -- Expected: 4

   -- If we used rem instead for backward movement:
   New_Index := (Current_Index - 1) rem BUFFER_SIZE;
   Put_Line ("(0 - 1) rem 5 = " & Integer'Image (New_Index)); -- Output: -1 (Incorrect for indexing)
end;

Using mod for robust circular buffer indexing, correctly handling both positive and 'negative' wraps. The rem operator would produce an invalid negative index.