注意Java除法运算的陷阱
【摘要】
除法运算谁不会啊,很多人不屑一顾,真是无知者无畏!
其实除法、求余运算有一些陷阱。一旦计算发生了问题,还很不好找。不好找的原因主要是问题的偶然性太强,如果你知道可能发生什么问题,你的代码就可以写得更安全。
数学除法规定,0不能做除数,因为会得到一个无穷大数据。
下面看看Java中如何处理这些特殊情况:
1、整数的除法: ...
下面看看Java中如何处理这些特殊情况:
1、整数的除法:
0做除数抛运行时异常;两整数商会做取整运算,Float或Double与一个整数做除法运算,则商位Float或者Double类型,例如:
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System.out.println("------------Int相关除法----------");
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System.out.println("12/10="+12/10);
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System.out.println("12f/10="+12f/10);
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System.out.println("12d/10="+12d/10);
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System.out.println("12/10f="+12/10d);
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System.out.println("12/10d="+12/10f);
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------------Int相关除法----------
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12/10=1
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12f/10=1.2
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12d/10=1.2
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12/10f=1.2
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12/10d=1.2
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package lavasoft.zerotest;
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/**
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* 浮点型数据的除法运算测试
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*
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* @author leizhimin 2009-12-21 9:00:37
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*/
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public class TestZero {
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public static void main(String[] args) {
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System.out.println("------------Double型----------");
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Double x1 = div(2.3, 0.0);
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Double x2 = div(2.3, -0.0);
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Double x3 = div(0.0, 0.0);
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Double x4 = div(0.0, -0.0);
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Double x5 = div(0.0, 0.1);
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Double x6 = div(0.0, -0.1);
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if (x1.isInfinite()) System.out.println("x1无穷大!");
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if (x2.isInfinite()) System.out.println("x2无穷大!");
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if (x3.isNaN()) System.out.println("x3非数字!");
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if (x4.isNaN()) System.out.println("x4非数字!");
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if (x1 == Double.POSITIVE_INFINITY) System.out.println("x1 = Double.POSITIVE_INFINITY");
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if (x2 == Double.NEGATIVE_INFINITY) System.out.println("x1 = Double.NEGATIVE_INFINITY");
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if (x3 == Double.NaN) System.out.println("x3 = Double.NaN");
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if (x4 == Double.NaN) System.out.println("x4 = -Double.NaN");
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System.out.println("---Float型---");
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Float y1 = div(2.3f, 0.0f);
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Float y2 = div(2.3f, -0.0f);
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Float y3 = div(0.0f, 0.0f);
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Float y4 = div(0.0f, -0.0f);
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Float y5 = div(0.0f, -0.1f);
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System.out.println("---比较测试---");
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Float a = 99999999999999999999999999999999999999f;
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Float b = 0.000000000000000000000000000000000000000000001f;
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Float t = a / b;
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System.out.println(t);
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System.out.println(Float.MAX_VALUE);
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if (t >= Float.MAX_VALUE) {
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System.out.println("a/b的商已经超过了Float的最大值了!");
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}
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}
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public static Double div(double a, double b) {
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double x = a / b;
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System.out.println(a + "/" + b + " = " + x);
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return x;
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}
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public static Float div(float a, float b) {
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float x = a / b;
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System.out.println(a + "/" + b + " = " + x);
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return x;
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}
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}
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运算输出:
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----------Double型----------
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2.3/0.0 = Infinity
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2.3/-0.0 = -Infinity
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0.0/0.0 = NaN
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0.0/-0.0 = NaN
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0.0/0.1 = 0.0
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0.0/-0.1 = -0.0
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x1无穷大!
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x2无穷大!
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x3非数字!
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x4非数字!
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x1 = Double.POSITIVE_INFINITY
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x1 = Double.NEGATIVE_INFINITY
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------------Float型----------
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2.3/0.0 = Infinity
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2.3/-0.0 = -Infinity
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0.0/0.0 = NaN
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0.0/-0.0 = NaN
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0.0/-0.1 = -0.0
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------------比较测试----------
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Infinity
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3.4028235E38
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a/b的商已经超过了Float的最大值了!
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Process finished with exit code 0
陷阱:
3、求余:和除法差不多。
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System.out.println(23%4);
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System.out.println(23%-4);
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System.out.println(-23%4);
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System.out.println(23f%4);
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System.out.println(23d%4);
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System.out.println(23%4f);
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System.out.println(23%4d);
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System.out.println(23f%0);
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System.out.println(23%0.0);
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System.out.println(0.0%0.0);
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运行结果:
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3
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3
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-3
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3.0
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3.0
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3.0
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3.0
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NaN
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NaN
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NaN
陷阱:
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public static final double POSITIVE_INFINITY = 1.0 / 0.0;
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public static final double NEGATIVE_INFINITY = -1.0 / 0.0;
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public static final double NaN = 0.0d / 0.0;
文章来源: panda1234lee.blog.csdn.net,作者:panda1234lee,版权归原作者所有,如需转载,请联系作者。
原文链接:panda1234lee.blog.csdn.net/article/details/8745979
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