在Java中找到最小的浮点数

Question

I really don't understand then,] as I not too good with maths. I've searched around but I can't really find anything to help me.

what I need to know is which one of these floating point numbers is the smallest

a. 28.268E-5f
b. 0.0000002763E4f
c. 282479.9E-9f
d. 0.2764E-2f

I'm not asking for the answer, if you don't want to give it to me, but more of a way for me to figure it out myself. In simple terms.

Answer 1

This is basic maths, not specific to Java. The "E(some number)" bit is short for "10 to the power of (some number)".

So 1E-5 is 1 x 10^-5 which is 0.00001 (1 divided by 10 5 times).

2E-5 would be 0.00002 and so on. Hopefully that gives you enough to figure out how to expand the numbers you have and see which is smallest.

Answer 2

The answer is c) 282479.9E-9f

For your clarity E stands for exponent that is 10 power the digit following it. so 1.0E-2 = 0.01

28.268E-5f     = 0.00028268 
0.0000002763E4 = 0.002763 
282479.9E-9    = 0.0002824799
0.2764E-2      = 0.0027464

Answer 3

probably the safest bet is to create a float[] containing your values, and use an array sorting algorithm to order them smallest to largest, from there-on-in, you can find the smallest, largest, medium values. Check this out. http://www.java-examples.com/java-sort-float-array-example