How to compare double and float primitive types in Java

Predict the output of the following two codes in Java:

 
// This program prints true
class Geeksforgeeks {
public static void main(String args[]) {
	float f = 5.25f;
	double d = 5.25
	System.out.println(f == d);
}
}

Output

true

 

// But this program prints false.
class Geeksforgeeks {
public static void main(String args[]) {
	float f = 5.1f;
	double d = 5.1;
	System.out.println(f == d);
}
}

Output

false

The output is true in the first example and false in second. We know that precisions of float and double are different. Size of mantissa for float is 24 and 53 for double.

Let us consider the first example of 5.25. Binary representation of integral part is 101 and binary representation of part of the dot is 0.01 (needs only two bits)

Let us consider the second example 5.1. The binary representation of an integral part is the same. But binary representation of 0.1 is 1/16 + 1/32 + 1/64 + 1/128 ….. and so on until we reach the end of mantissa or sum becomes more than 0.1. In this case, we reach the end of mantissa and therefore the value of 5.1 becomes different in float and double.

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