Floating Point Operations & Associativity in C, C++ and Java

Do Floating point operations follow property of associativity? In other words, do we always get the same results for expressions “(A + B) + C” and “A + (B + C)” One may expect that floating numbers to follow the rule of associativity in programming languages as they are associative mathematically. However, this is not true in all cases. Consider below C/C++ program. 

C

// C/C++ program to demonstrate that floating point
// addition may not be associative.
#include <stdio.h>

int main()
{
    // A and B have same values but opposite signs
    float A = -500000000;
    float B =  500000000;

    float C = 1;

    printf("A + (B + C) is equal to %f \n", A + (B + C));
    printf("(A + B) + C is equal to %f", (A + B) + C);

    return 0;
}

Output:

A + (B + C) is equal to 0.000000 
(A + B) + C is equal to 1.000000

It is evident from the above-given output that the floating-point arithmetic may not follow the law of associativity in every case. This is due to the format in which the floating-point numbers are stored and represented, it rounds off the numbers during calculations, hence, the associative laws of algebra do not necessarily hold for floating-point numbers. In this case,

Explanation for above output:

A + (B + C):
(B + C) = 500000000.0 + 1.0
        = 500000000.0 
(rounded off during floating point arithmetic)

A + (B + C) = -500000000.0 + 500000000.0 
            =  0.000000

(A + B) + C:
(A + B) = -500000000.0 + 500000000.0 
        = 0.000000

(A + B) + C = 0.000000 + 1 
            = 1.000000

How about Java? We get the same results in Java as Java also uses similar representation for floating-point numbers. 

Java

// Java program to demonstrate that floating-point
// addition may not be associative
import java.io.*;

class Main
{
    public static void main (String[] args)
    {
        // A and B have same values but opposite signs
        float A = -500000000;
        float B =  500000000;

        float C = 1;

        System.out.println("A + (B + C) is equal to " +
                          (A + (B + C)));
        System.out.println("(A + B) + C is equal to " +
                          ((A + B) + C));
    }
}

Output:

A + (B + C) is equal to 0.000000 
(A + B) + C is equal to 1.000000

How about integers? Now let’s try the same calculations when the data type is integer. Here is a piece of code for your observation: 

C++

#include <iostream>
using namespace std;

int main()
{
   // A and B have same values but opposite signs
   int A = -500000000;
   int B =  500000000;

   int C = 1;

   cout << " A + (B + C) is equal to " << A + (B + C) << endl;
   cout << "(A + B) + C is equal to " << (A + B) + C << endl;

   return 0;
}

// This code is contributed by sarajadhav12052009

C

#include <stdio.h>

int main()
{
   // A and B have same values but opposite signs
   int A = -500000000;
   int B =  500000000;

   int C = 1;

   printf(" A + (B + C) is equal to %d \n", A + (B + C));
   printf("(A + B) + C is equal to %d", (A + B) + C);

   return 0;
}

Java

import java.io.*;

class Main
{
    public static void main (String[] args)
    {
        // A and B have same values but opposite signs
        int A = -500000000;
        int B =  500000000;

        int C = 1;

        System.out.println("A + (B + C) is equal to " +
                          (A + (B + C)));
        System.out.println("(A + B) + C is equal to " +
                          ((A + B) + C));
    }
}

// This code is contributed by sarajadhav12052009

Output

 A + (B + C) is equal to 1
(A + B) + C is equal to 1

Time Complexity: O(1).
Auxiliary Space: O(1).

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