为什么Java使用这个长二进制数来使用Long.parseLong（String s，int radix）会冒犯？

Question

I have the following code:

Why does Java think that this is not a valid long .

@Test
public void testOffendingBinaryString() {
  String offendingString = "1000000000000000000010101000000000000000000000000000000000000000";
  assertEquals(64, offendingString.length());
  Long.parseLong(offendingString, 2);
}

Answer 1

Because it's out of range for the valid value of a long. The string:

"-111111111111111111101011000000000000000000000000000000000000000"

should work just fine. You cannot specify a negative number directly in 2s complement binary notation . And if you're trying to specify a positive number, that number is too big to fit into a Java long, which is a signed 64-bit value expressed in 2s complement (which means it's basically 63 bits + a sign bit (it's a little more complicated than that, read the page on 2s complement)).

Answer 2

这是Java中的一个错误，请参阅http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4215269 ，他们将其修复为1.8

Answer 3

The long is stored in 2's compliment, but the parse method expects straight binary. So, the max number of digits in the String can be 63.

When using regular binary, there is no sign bit. You can pass it a String with length 63 and precede it with a - if you want a negative long .

Answer 4

Java long type:

8 bytes signed (two's complement). Ranges from -9,223,372,036,854,775,808 to +9,223,372,036,854,775,807.

Answer 5

字符串的大小太长，请检查此问题

Answer 6

Internally numbers are represented using the Two's complement . That means there are only 63 bits, for the number and one bit to represent positive or negative numbers. Since Long.parseLong() is not ment to parse numbers in Two's complement your String is a positive number with 64 bits, that exceeds the maximum positive value for a long .