Exception in thread "main" java.lang.NumberFormatException: For input string: "9000000000000000" under radix 16

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英文:

Exception in thread "main" java.lang.NumberFormatException: For input string: "9000000000000000" under radix 16

问题

我尝试运行这段代码,但出现了错误。

System.out.println(Long.parseLong("9000000000000000", 16));

众所周知,long的最小值是-9,223,372,036,854,775,808,而0x9000000000000000对应的是-8,070,450,532,247,928,832,为什么会出现错误呢?

英文:

I try to run this code but it occur error.

System.out.println(Long.parseLong("9000000000000000", 16));

As we know that the minimum number of long is -9,223,372,036,854,775,808 and 0x9000000000000000 is -8,070,450,532,247,928,832 why it will occur error?

答案1

得分: 10

9000000000000000在16进制中表示的是一个正数,因为没有符号。由于long是有符号的,它能够表示的最大数字是0x7FFF_FFFF_FFFF_FFFF。所以你的数值太大了。

如果你想要表示-8,070,450,532,247,928,832,可以使用parseUnsignedLong()函数:

System.out.println(Long.parseUnsignedLong("9000000000000000", 16));

输出结果:

> -8070450532247928832

现在可以接受的值范围扩展到了0xFFFF_FFFF_FFFF_FFFF。

英文:

9000000000000000 base 16 is a positive number since there is no sign. Since a long is signed, the greatest number that it can hold is 0x7FFF_FFFF_FFFF_FFFF. So yours is too great.

If you want -8,070,450,532,247,928,832, use parseUnsignedLong():

		System.out.println(Long.parseUnsignedLong("9000000000000000", 16));

Output:

> -8070450532247928832

Now values up to 0xFFFF_FFFF_FFFF_FFFF are accepted.

答案2

得分: 2

关于 Long#parseLong(String,int) 的解释:

如果出现以下任何情况,将抛出 NumberFormatException 异常:

  • 第一个参数为 null 或长度为零的字符串。
  • 基数小于 Character.MIN_RADIX 或大于 Character.MAX_RADIX。
  • 字符串中的任何字符不是指定基数的数字,但如果字符串长度大于1,则第一个字符可以是减号 '-' ('\u002d') 或加号 '+' ('\u002B')。
  • 字符串表示的值不是 long 类型的值。

示例:

  • parseLong("0", 10) 返回 0L
  • parseLong("473", 10) 返回 473L
  • parseLong("+42", 10) 返回 42L
  • parseLong("-0", 10) 返回 0L
  • parseLong("-FF", 16) 返回 -255L
  • parseLong("1100110", 2) 返回 102L
  • parseLong("99", 8) 抛出 NumberFormatException
  • parseLong("Hazelnut", 10) 抛出 NumberFormatException
  • parseLong("Hazelnut", 36) 返回 1356099454469L

使用基数 16 解析的十进制值为 10376293541461622784,大于 Long.MAX_VALUE(9223372036854775807),违反以下条件:

  • 字符串表示的值不是 long 类型的值

因此会抛出 NumberFormatException

使用基数 16 解析的十进制值为 10376293541461622784,大于 Long.MAX_VALUE(9223372036854775807),违反以下条件:

字符串表示的值不是 long 类型的值

因此会抛出 NumberFormatException 异常。

import java.math.BigInteger;

public class ParseLong {
    public static void main(String[] args) {
        System.out.println("Max Long value :" + Long.MAX_VALUE);
        System.out.println("Min Long value :" + Long.MIN_VALUE);
        System.out.println("Value without overflow " + new BigInteger("9000000000000000", 16));
        System.out.println("Value by parseUnsigned " + Long.parseUnsignedLong("9000000000000000", 16));
        System.out.println("Value by literal " + 0x9000000000000000L);
    }
}
英文:

Referring to Long#parseLong(String,int)

> An exception of type NumberFormatException is thrown if any of the
> following situations occurs:
>
> - The first argument is null or is a string of length zero.
> - The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX.
> - Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\u002d') or plus sign '+' ('\u002B') provided that the string is longer than length 1.
> - The value represented by the string is not a value of type long.
>
> Examples:
>   parseLong("0", 10) returns 0L
>   parseLong("473", 10) returns 473L
>   parseLong("+42", 10) returns 42L
>   parseLong("-0", 10) returns 0L
>   parseLong("-FF", 16) returns -255L
>   parseLong("1100110", 2) returns 102L
>   parseLong("99", 8) throws a NumberFormatException
>   parseLong("Hazelnut", 10) throws a NumberFormatException
>   parseLong("Hazelnut", 36) returns 1356099454469L

The decimal value parsed using radix 16 is 10376293541461622784 which is greater than Long.MAX_VALUE(9223372036854775807), violate following condition:

>The value represented by the string is not a value of type long

hence throwing NumberFormatException.

import java.math.BigInteger;

public class ParseLong {
	public static void main(String[] args) {
		System.out.println("Max Long value :" + Long.MAX_VALUE);
		System.out.println("Min Long value :" + Long.MIN_VALUE);
		System.out.println("Value without overflow " + new BigInteger("9000000000000000", 16));
		System.out.println("Value by parseUnsigned " + Long.parseUnsignedLong("9000000000000000", 16));
		System.out.println("Value by literal " + 0x9000000000000000L);
	}
}

答案3

得分: 1

Long.parseLong()不会像算术运算一样“溢出”到负数 - 它不是解析位表示,而是整数的数字。

十六进制中最大的long7FFFFFFFFFFFFFFF;您的值比这还要大。

以十进制进行比较:

十六进制          十进制
7FFFFFFFFFFFFFFF  9,223,372,036,854,775,807
9000000000000000  10,376,293,541,461,622,784
英文:

Long.parseLong() does not "overflow" around to a negative number like arithmetic does - it's not a parsing a bit representation but rather the digits of a whole number.

The largest long in base 16 is 7FFFFFFFFFFFFFFF; your value is bigger than that.

For comparison in decimal:

Base 16            Decimal
7FFFFFFFFFFFFFFF    9,223,372,036,854,775,807
9000000000000000   10,376,293,541,461,622,784

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  • 本文由 发表于 2020年9月10日 12:29:03
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