Java,Long.parse二进制字符串

Question

为什么这段代码会抛出NumberFormatException:

String binStr = "1000000000000000000000000000000000000000000000000000000000000000";
System.out.println(binStr.length());// =  64
System.out.println(Long.parseLong(binStr, 2));

Answer 1

1000000000000000000000000000000000000000000000000000000000000000大于Long.MAX_VALUE.

请参阅/sf/answers/622227861/

考虑BigInteger(String val, int radix)改用.

编辑:

好的,这对我来说是新的.它似乎Integer.parseInt(binaryIntegerString, 2)并将Long.parseLong(binaryLongString, 2)二进制解析为符号量,而不是2的补码.

Answer 2

因为它超出了范围. 1000...000是2 ⁶³,但Long最多只有2 ⁶³ - 1.