Android除数(带小数部分)

问题描述

如何检查数字是否可以被特定数字整除或两个数字都不都是小数.小数点后两位数字的值.我曾经尝试过(((dn / size) % 1) == 0)，但在某些情况下不能提供适当的输出.我该如何解决.在这里我放一些示例值，例如double dn = 369.35,369.55.370.55和大小可能是0.05,0.10,0.5等...

How can i check number is divisible by particular number or not both numbers are decimal. with decimal point value of two digits.i had tried with below(((dn / size) % 1) == 0) but in some cases it not provide proper out put.how can i resolve it. here i put some example values likedouble dn = 369.35,369.55.370.55 and size may be 0.05,0.10,0.5 etc...

 if(((dn / size) % 1) == 0)    {
       Log.d(TAG,"OK");
    } else {
       Log.d(TAG,"OK");
    }

请帮我把它短路.

推荐答案

(dn / size) % 1 == 0虽然合理，但将遭受以二进制浮点运算为中心的陷阱.

(dn / size) % 1 == 0 whilst plausible, will suffer from pitfalls centred around binary floating point arithmetic.

如果您的数字始终不超过两位小数，那么最简单的方法是乘以100并按100 * y除100 * x的事实进行扩展，如果y除以x.换句话说:

If your numbers always have no more then two decimal places then the easiest thing to do is scale up by multiplying by 100 and rely on the fact that 100 * y divides 100 * x if y divides x. In other words:

if (Math.round(100 * dn) % Math.round(100 * size) == 0){
    Log.d(TAG,"Divisible");
} else {
    Log.d(TAG,"Not divisible");
}

这消除了二进制浮点算术的所有问题.我再次使用Math.round而不是故意使用截断符来消除浮点周围的问题，并且依靠我认为该平台的怪癖，因为round返回整数类型.

This obviates any issues with binary floating point arithmetic. I've used Math.round rather than a truncation deliberately, again to obviate issues around floating point and am relying on what is in my opinion a quirk of this platform in that round returns an integral type.

进一步阅读:浮点数学运算是否被破坏?

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