How float with 'e' will work in a variable declaration

Question

public class TypeConversion3 {

    public static void main(String[] args) {
        float f = 12e-1F;
        final long l = 12L;
        f = f + l;
        System.out.println(f); //prints 13.2
    }
}

No idea how it prints the value 13.2. I some where read that e is 2.718 but what is this float f = 12e-1F; representing ?

score 4 · Accepted Answer · answered May 01 '15 at 08:35

4

The "e" is not the natural logarithm base, but short for "exponent". The number following the "e" is a power of ten, so 12e-1F is 1.2.

answered May 01 '15 at 08:35

Patricia Shanahan

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score 3 · Answer 2 · answered May 01 '15 at 08:35

3

12e-1 is scientific notation and means 12 * 10^-1 which is 1.2

answered May 01 '15 at 08:35

Raniz

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YoYo · Answer 3 · 2015-05-01T08:57:30.963

2

The e stands for 10^, or power of 10.

12E-1F = 12.0 * 10 ^ -1 = 12.0 / 10 = 1.2

It is part of the syntax as defined in the Java Language Specification.

The Base for a Natural Logarithm (e) is available as Math.E (Eulers Number), or to use it as the base for an exponent, simple use Math.exp(-1). Note however that parameters and result types are double not float.

edited May 01 '15 at 08:57

answered May 01 '15 at 08:39

YoYo

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so -1 is the power in this case ? – kittu May 01 '15 at 08:44
1

10 raised to the power of (-1) yes. – YoYo May 01 '15 at 08:59

score 2 · Answer 4 · answered May 01 '15 at 09:09

2

You are confusing E notation with Euler's constant which is approx 2.71828.

Java number literals use e as E notation thus e in 12e-1F means float number 12*10^-1 which is 1.2.

answered May 01 '15 at 09:09

Piotrek Bzdyl

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How float with 'e' will work in a variable declaration

4 Answers4