将一个浮点数整数到java中的下一个整数值 [英] round a floating-point number to the next integer value in java

问题描述

我怎样才能在Java中将浮点数加到下一个整数值？假设

2.1 - > 3

3.001 - > 4 p>

4.5 - > 5

7.9 - > 8

解决方案

你应该看看Java的数学包中的上限四舍五入：Math.ceil

---

编辑：添加Math.ceil的javadoc。这可能是值得读数学的所有方法。

http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#ceil%28double%29

public static double ceil（double a）

返回最小（最接近负无穷大）double值，即
大于或等于参数，等于
数学整数。特殊情况：

• 如果参数值已经等于一个数学整数，那么结果和参数一样。
>

• 如果参数是NaN或无穷大或正零或负零，则结果与参数相同。
  
• 如果参数值较小

请注意，数学的值.ceil（x）正是
-Math.floor（-x）的值。

how can i round up a floating point number to the next integer value in Java? Suppose

2.1 -->3

3.001 -->4

4.5 -->5

7.9 -->8

解决方案

You should look at ceiling rounding up in java's math packages: Math.ceil

---

EDIT: Added the javadoc for Math.ceil. It may be worth reading all the method in Math.

http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#ceil%28double%29

public static double ceil(double a)

Returns the smallest (closest to negative infinity) double value that is greater than or equal to the argument and is equal to a mathematical integer. Special cases:

• If the argument value is already equal to a mathematical integer, then the result is the same as the argument.
• If the argument is NaN or an infinity or positive zero or negative zero, then the result is the same as the argument.
• If the argument value is less than zero but greater than -1.0, then the result is negative zero.

Note that the value of Math.ceil(x) is exactly the value of -Math.floor(-x).

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