BigDecimal.valueOf（float）如何工作？

Question

I've already looked at this asnwer .

I was a little confused by the behaviour of BigDecimal.valueOf(float) . Here's the example:

System.out.println(BigDecimal.valueOf(20.2f)); //prints 20.200000762939453

Therefore

float f = BigDecimal.valueOf(20.2f)
            .setScale(2, RoundingMode.UP)
            .floatValue();
System.out.println(f); //prints 20.21

Which is defenitely not the behaviour I expected. Is there a way to round up float correctly avoiding such errors?

Answer 1

For the description of float and how it utilizes the allocated bits see How many significant digits have floats and doubles in java?

BigDecimal works fine and -- as expected -- keeps all digits it received, but it can't guess the precision of the given argument. Therefore:

    float f = 20.20f;
    System.out.println("Float: "+f);
    System.out.println("BigDecimal.valueOf: "+BigDecimal.valueOf(f));
    System.out.println("BigDecimal from value: "+new BigDecimal(f, MathContext.DECIMAL32));
    System.out.println("BigDecimal from string: "+new BigDecimal(""+f));;

Prints:

Float: 20.2
BigDecimal.valueOf: 20.200000762939453
BigDecimal from value: 20.20000
BigDecimal from string: 20.2