What's the right way to parseFloat in Java

2020-03-12 04:58发布

I notice some issues with the Java float precision

       Float.parseFloat("0.0065") - 0.001  // 0.0055000000134110451
       new Float("0.027") - 0.001          // 0.02600000000700354575
       Float.valueOf("0.074") - 0.001      // 0.07399999999999999999

I not only have a problem with Float but also with Double.

Can someone explain what is happening behind the scenes, and how can we get an accurate number? What would be the right way to handle this when dealing with these issues?

7条回答
叼着烟拽天下
2楼-- · 2020-03-12 05:34

Floating point cannot accurately represent decimal numbers. If you need an accurate representation of a number in Java, you should use the java.math.BigDecimal class:

BigDecimal d = new BigDecimal("0.0065");
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再贱就再见
3楼-- · 2020-03-12 05:42

The problem is simply that float has finite precision; it cannot represent 0.0065 exactly. (The same is true of double, of course: it has greater precision, but still finite.)

A further problem, which makes the above problem more obvious, is that 0.001 is a double rather than a float, so your float is getting promoted to a double to perform the subtraction, and of course at that point the system has no way to recover the missing precision that a double could have represented to begin with. To address that, you would write:

float f = Float.parseFloat("0.0065") - 0.001f;

using 0.001f instead of 0.001.

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▲ chillily
4楼-- · 2020-03-12 05:46

I would convert your float to a string and then use BigDecimal.

This link explains it well

new BigDecimal(String.valueOf(yourDoubleValue));

Dont use the BigDecimal double constructor though as you will still get errors

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我只想做你的唯一
5楼-- · 2020-03-12 05:49

See What Every Computer Scientist Should Know About Floating-Point Arithmetic. Your results look correct to me.

If you don't like how floating-point numbers work, try something like BigDecimal instead.

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姐就是有狂的资本
6楼-- · 2020-03-12 05:51

Long story short if you require arbitrary precision use BigDecimal not float or double. You will see all sorts of rounding issues of this nature using float.

As an aside be very careful not to use the float/double constructor of BigDecimal because it will have the same issue. Use the String constructor instead.

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【Aperson】
7楼-- · 2020-03-12 05:52

From the Java Tutorials page on Primitive Data Types:

A floating-point literal is of type float if it ends with the letter F or f; otherwise its type is double and it can optionally end with the letter D or d.

So I think your literals (0.001) are doubles and you're subtracting doubles from floats.

Try this instead:

System.out.println((0.0065F - 0.001D)); // 0.005500000134110451
System.out.println((0.0065F - 0.001F)); // 0.0055

... and you'll get:

0.005500000134110451
0.0055

So add F suffixes to your literals and you should get better results:

Float.parseFloat("0.0065") - 0.001F
new Float("0.027") - 0.001F
Float.valueOf("0.074") - 0.001F
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