When should I use the “strictfp” keyword in java?

2019-01-02 19:31发布

I've looked up what this does, but does anyone actually have an example of when you would use the strictfp keyword in Java? Has anyone actually found a use for this?

Would there be any side-effects of just putting it on all my floating point operations?

8条回答
明月照影归
2楼-- · 2019-01-02 19:57

Wikipedia actually has a good article about this topic here, with a link to the Java specification.

Reading between the lines, the implication is that if you don't specify strictfp, then the JVM and JIT compiler have license to compute your floating-point calculations however they want. In the interest of speed, they will most likely delegate the computation to your processor. With strictfp on, the computations have to conform to IEEE 754 arithmetic standards, which, in practice, probably means that the JVM will do the computation.

So why would you want to use strictfp? One scenario I can see is in a distributed application (or multiplayer game) where all floating-point calculations need to be deterministic no matter what the underlying hardware or CPU is. What's the trade-off? Most likely execution time.

查看更多
旧时光的记忆
3楼-- · 2019-01-02 20:00

Here are several references:

  • Using strictfp (JDC Tech Tip)
  • jGuru: What is the strictfp modifier for? When would I consider using it?

    Basically, what it all boils down to is whether or not you care that the results of floating-point expressions in your code are fast or predictable. For example, if you need the answers that your code comes up with which uses floating-point values to be consistent across multiple platforms then use strictfp.

  • strictfp - Java Glossary

    Floating point hardware calculates with more precision, and with a greater range of values than the Java specification requires. It would be confusing if some platforms gave more precision than others. When you use the strictfp modifier on a method or class, the compiler generates code that adheres strictly to the Java spec for identical results on all platforms. Without strictfp, is it is slightly laxer, but not so lax as to use the guard bits in the Pentium to give 80 bits of precision.

  • And finally the actual Java Language Specification, §15.4 FP-strict Expressions:

    Within an FP-strict expression, all intermediate values must be elements of the float value set or the double value set, implying that the results of all FP-strict expressions must be those predicted by IEEE 754 arithmetic on operands represented using single and double formats. Within an expression that is not FP-strict, some leeway is granted for an implementation to use an extended exponent range to represent intermediate results; the net effect, roughly speaking, is that a calculation might produce "the correct answer" in situations where exclusive use of the float value set or double value set might result in overflow or underflow.

I've never personally had a use for it, though.

查看更多
呛了眼睛熬了心
4楼-- · 2019-01-02 20:00

May below example help in understanding this more clear : In java whenever we are using looking for precise information for any operation e.g. if we do double num1 = 10e+102; double num2 = 8e+10 ; result = num1+ num2;

        The output will be so long and not precise, becasue it is precissed by the hardware e.g JVM and JIT has the license 
        as long as we dont have specify it Strictfp

Marking it Strictfp will make the result Uniform on every hardware and platform, because its precised value will be same
One scenario I can see is in a distributed application (or multiplayer game) where all floating-point calculations need to 
be deterministic no matter what the underlying hardware or CPU is.
查看更多
琉璃瓶的回忆
5楼-- · 2019-01-02 20:06

It all began with a story,

When java was being developed by James Gosling, Herbert and rest of his team. They had this crazy thing in mind called platform independency. They wanted to make oak(Java) so much better that it would run exactly same on any machine having different instruction set, even running different operating systems. But, there was a problem with decimal point numbers also known as floating point and double in programming languages. Some machines were built targeting efficiency while rest were targeting accuracy. So, the later(more accurate) machines had size of floating point as 80 bits while the former(more efficient/faster) machines had 64 bit doubles. But, this was against there core idea of building a platform independent language. Also, this might lead to loss of precision/data when a code is built on some machine(having double of 64 bit size) and run on another kind of machine(having double of 80 bit size).

Up-Sizing can be tolerated but Down-Sizing can't be. So, they came across a concept of strictfp i.e. strict floating point. If you use this keyword with a class/function then its floating point and doubles have a consistent size over any machine. i.e. 32/64 -bit respectively.

查看更多
梦醉为红颜
6楼-- · 2019-01-02 20:09

Strictfp ensures that you get exactly the same results from your floating point calculations on every platform. If you don't use strictfp, the JVM implementation is free to use extra precision where available.

From the JLS:

Within an FP-strict expression, all intermediate values must be elements of the float value set or the double value set, implying that the results of all FP-strict expressions must be those predicted by IEEE 754 arithmetic on operands represented using single and double formats. Within an expression that is not FP-strict, some leeway is granted for an implementation to use an extended exponent range to represent intermediate results; the net effect, roughly speaking, is that a calculation might produce "the correct answer" in situations where exclusive use of the float value set or double value set might result in overflow or underflow.

In other words, it's about making sure that Write-Once-Run-Anywhere actually means Write-Once-Get-Equally-Wrong-Results-Everywhere.

With strictfp your results are portable, without it they are more likely to be accurate.

查看更多
其实,你不懂
7楼-- · 2019-01-02 20:09

As the other answers mentioned it cause the intermediate floating point results to conform to the IEEE specification. In particular x86 processors can store intermediate results with different precision from the IEEE spec. The situation gets more complicated when the JIT optimizes a particular computation; the order the instructions could be different each time resulting in slightly different rounding.

The overhead incurred by strictfp likely to be very processor and JIT dependent. This wikipedia article on SSE2 seems to have some insight into the problem. So if the JIT can generate SSE instructions to perform a calculation it seems that strictfp will not have any overhead.

In my current project there are a few places where I use strictfp. There is a point where potential cosmic rays need to be removed from pixel values. If some outside researcher has the the same pixel value and cosmic ray in front them they should get the same resulting value as our software.

查看更多
登录 后发表回答