Float vs double Math Java

Question

Is there a precision difference between the following (assuming the value of a and b can be represented without loss of precision in a float).

With floats:

float a;
float b;
double result = 1 + a*b;

With doubles:

double a;
double b;
double result = 1 + a*b;

Answer 1

Simple example:

float a = 16777217;  // Largest int exactly representable in a float.
float b = 16777217;
System.out.println((double)(1 + a*b));

double c = 16777217;
double d = 16777217;
System.out.println(1 + c*d);

Output (Ideone):

2.81474976710656E14
2.8147501026509E14

So yes, there is a loss of precision using float.

Answer 2

There is a loss of precision in

float a;
float b;
double result = 1 + a*b;

• the float representation.
• the product of a and b which will also be a float. Note: a * b is a float
• the addition on 1 could result in a loss of precision.

To should that a * b can lose more precision

for (int i = 1; i < 100; i += 2) {
    float a = i;
    float b = 1.0f / i;
    if ((double) a * b != a * b && a * b != 1)
        System.out.println(i + " " + (double) a * b + " " + a * b);
}

prints

41 0.999999962747097 0.99999994
47 0.9999999683350325 0.99999994
55 0.9999999683350325 0.99999994
61 0.9999999441206455 0.99999994
83 0.999999962747097 0.99999994
97 0.999999969266355 0.99999994

note: it could also happen recover precision lost and get the right answer after b has lost precision

for (int i = 1; i < 20; i += 2) {
    float a = i;
    float b = 1.0f / i;
    if (b != 1.0 / i && a * b == 1)
        System.out.println(i + " " + (double) a * b + " " + a * b);
}

prints

3 1.0000000298023224 1.0
5 1.0000000149011612 1.0
7 1.0000000447034836 1.0
9 1.0000000074505806 1.0
11 1.0000000298023224 1.0
13 1.000000037252903 1.0
15 1.0000000521540642 1.0
17 1.0000000037252903 1.0
19 1.0000000074505806 1.0