Java BigInteger sqrt() 方法

7 Jan 2025 | 1 分钟阅读

Java BigInteger 类的 sqrt() 方法用于查找 BigInteger 值的平方根。

语法

参数

不适用

返回值

此方法返回此 BigInteger 值的整数平方根。

Exception

ArithmeticException - 如果此 BigInteger 为负数，则会抛出此异常。

示例 1

import java.math.BigInteger;
public class BigIntegerSqrtExample1 {	
 public static void main(String[] args) {		
   //create 2 BigInteger objects 
   BigInteger squareRoot,result;
   // initialize bigInteger objects
   squareRoot =new BigInteger("625");
   //perform sqrt operation on squareRoot
   result= squareRoot.sqrt();
   System.out.println("Square root of BigInteger "+squareRoot+" is "+result);
  }
}

输出

Square root of BigInteger 625 is 25

示例 2

import java.math.BigInteger;
public class BigIntegerSqrtExample2 {	
 public static void main(String[] args) {		
   //create 2 BigInteger objects 
   BigInteger squareRoot,result;
   // initialize bigInteger objects
   squareRoot =new BigInteger("1000");
   //perform sqrt operation on squareRoot
   result= squareRoot.sqrt();
   System.out.println("Square root of BigInteger "+squareRoot+" is "+result);
  }
}

输出

Square root of BigInteger 1000 is 31

示例 3

import java.math.BigInteger;
public class BigIntegerSqrtExample3 {	
 public static void main(String[] args) {		
   //create 2 BigInteger objects 
   BigInteger squareRoot,result;
   // initialize bigInteger objects
   squareRoot=new BigInteger("-1000");
   //perform sqrt operation on squareRoot
   result= squareRoot.sqrt();
   System.out.println("Square root of BigInteger "+squareRoot+" is "+result);
  }
}

输出

Exception in thread "main" java.lang.ArithmeticException: Negative BigInteger
	at java.base/java.math.BigInteger.sqrt(Unknown Source)
	at BigIntegerSqrtExample3.main(BigIntegerSqrtExample3.java:9)

下一个主题Java Boolean 类

Java BigInteger sqrt() 方法

语法

参数

返回值

Exception

示例 1

示例 2

示例 3

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Java BigInteger sqrt() 方法

语法

参数

返回值

Exception

示例 1

示例 2

示例 3

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