math round 用法（Math.round5.5等于多少）

【死记硬背】Math类提供了三个与取整有关的方法：ceil、floor和round，我来为大家讲解一下关于math round 用法?跟着小编一起来看一看吧!

math round 用法

【死记硬背】

Math类提供了三个与取整有关的方法：ceil、floor和round。

ceil：英文意思为天花板，表示向上取整，Math.ceil(11.3)的结果为12.0，Math.ceil(-11.3)的结果是-11.0；

floor：英文意思为地板，表示向下取整，Math.floor(11.6)的结果为11.0，Math.floor(-11.6)的结果是-12.0;

round：英文意思为原型的，表示四舍五入，Math.round(5.5)的结果为6，Math.round(-5.5)的结果为-5，

【答案解析】

示例代码如下：

public class Test{ public static void main(String[] args) { double ceil1 = Math.ceil(11.3); double ceil2 = Math.ceil(-11.3); System.out.println("Math.ceil(11.3) = " ceil1); System.out.println("Math.ceil(-11.3) = " ceil2); double floor1 = Math.floor(11.6); double floor2 = Math.floor(-11.6); System.out.println("Math.floor(11.6) = " floor1); System.out.println("Math.floor(-11.6) = " floor2); long round1 = Math.round(5.5); long round2 = Math.round(-5.5); System.out.println("Math.round(5.5) = " round1); System.out.println("Math.round(-5.5) = " round2); } } // 输出内容如下 Math.ceil(11.3) = 12.0 Math.ceil(-11.3) = -11.0 Math.floor(11.6) = 11.0 Math.floor(-11.6) = -12.0 Math.round(5.5) = 6 Math.round(-5.5) = -5

Math.ceil的源码解析：

// Math.ceil的方法： public static double ceil(double a) { return StrictMath.ceil(a); } // StrictMath.ceil的方法： public static double ceil(double a) { return floorOrCeil(a, -0.0, 1.0, 1.0); }

Math.floor的源码解析：

// Math.floor的方法： public static double floor(double a) { return StrictMath.floor(a); } // StrictMath.floor的方法 public static double floor(double a) { return floorOrCeil(a, -1.0, 0.0, -1.0); }

Math.ceil和Math.floor最终共用的方法：

// floorOrCeil的方法 private static double floorOrCeil(double a, double negativeBoundary, double positiveBoundary, double sign) { int exponent = Math.getExponent(a); if (exponent < 0) { /* * Absolute value of argument is less than 1. * floorOrceil(-0.0) => -0.0 * floorOrceil( 0.0) => 0.0 */ return ((a == 0.0) ? a : ( (a < 0.0) ? negativeBoundary : positiveBoundary) ); } else if (exponent >= 52) { /* * Infinity, NaN, or a value so large it must be integral. */ return a; } // Else the argument is either an integral value already XOR it // has to be rounded to one. assert exponent >= 0 && exponent <= 51; long doppel = Double.doubleToRawLongBits(a); long mask = DoubleConsts.SIGNIF_BIT_MASK >> exponent; if ( (mask & doppel) == 0L ) return a; // integral value else { double result = Double.longBitsToDouble(doppel & (~mask)); if (sign*a > 0.0) result = result sign; return result; } }

Math.round的源码解析：

public static long round(double a) { long longBits = Double.doubleToRawLongBits(a); long biasedExp = (longBits & DoubleConsts.EXP_BIT_MASK) >> (DoubleConsts.SIGNIFICAND_WIDTH - 1); long shift = (DoubleConsts.SIGNIFICAND_WIDTH - 2 DoubleConsts.EXP_BIAS) - biasedExp; if ((shift & -64) == 0) { // shift >= 0 && shift < 64 // a is a finite number such that pow(2,-64) <= ulp(a) < 1 long r = ((longBits & DoubleConsts.SIGNIF_BIT_MASK) | (DoubleConsts.SIGNIF_BIT_MASK 1)); if (longBits < 0) { r = -r; } // In the comments below each Java expression evaluates to the value // the corresponding mathematical expression: // (r) evaluates to a / ulp(a) // (r >> shift) evaluates to floor(a * 2) // ((r >> shift) 1) evaluates to floor((a 1/2) * 2) // (((r >> shift) 1) >> 1) evaluates to floor(a 1/2) return ((r >> shift) 1) >> 1; } else { // a is either // - a finite number with abs(a) < exp(2,DoubleConsts.SIGNIFICAND_WIDTH-64) < 1/2 // - a finite number with ulp(a) >= 1 and hence a is a mathematical integer // - an infinity or NaN return (long) a; } }

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