Java，Long.parse 二进制字符串
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Java, Long.parse binary String
提问by Mickey Tin
Why does this code throw a NumberFormatException:
为什么这段代码会抛出一个NumberFormatException：
String binStr = "1000000000000000000000000000000000000000000000000000000000000000";
System.out.println(binStr.length());// = 64
System.out.println(Long.parseLong(binStr, 2));
采纳答案by Eng.Fouad
1000000000000000000000000000000000000000000000000000000000000000is larger than Long.MAX_VALUE.
1000000000000000000000000000000000000000000000000000000000000000大于Long.MAX_VALUE。
See https://stackoverflow.com/a/8888969/597657
见https://stackoverflow.com/a/8888969/597657
Consider using BigInteger(String val, int radix)instead.
考虑BigInteger(String val, int radix)改用。
EDIT:
编辑：
OK, this is new for me. It appears that Integer.parseInt(binaryIntegerString, 2)and Long.parseLong(binaryLongString, 2)parse binary as sign-magnitude not as a 2's-complement.
好的，这对我来说是新的。看来，Integer.parseInt(binaryIntegerString, 2)和Long.parseLong(binaryLongString, 2)解析二进制的符号-幅度不作为二进制补码。
回答by Oliver Charlesworth
Because it's out of range. 1000...000is 263, but Longonly goes up to 263- 1.
因为超出范围了。 1000...000是 2 63，但Long只会上升到 2 63- 1。
回答by jlordo
This is the same for all of Long, Integer, Shortand Byte. I'll explain with a Byteexample because it's readable:
这是所有的相同Long，Integer，Short和Byte。我会用一个Byte例子来解释，因为它是可读的：
System.out.println(Byte.MIN_VALUE); // -128
System.out.println(Byte.MAX_VALUE); // 127
String positive = "1000000"; // 8 binary digits, +128
String negative = "-1000000"; // 8 binary digits, -128
String plus = "+1000000"; // 8 binary digits, +128
Byte.parseByte(positive, 2); //will fail because it's bigger than Byte.MAX_VALUE
Byte.parseByte(negative, 2); //won't fail. It will return Byte.MIN_VALUE
Byte.parseByte(plus, 2); //will fail because its bigger than Byte.MAX_VALUE
The digits are interpreted unsigned, no matter what radix is provided. If you want a negative value, you have to have the minus sign at the beginning of the String. JavaDocsays:
无论提供什么基数，数字都被解释为无符号。如果你想要一个负值，你必须在字符串的开头有一个减号。JavaDoc说：
Parses the string argument as a signed long in the radix specified by the second argument. The characters in the string must all be digits of the specified radix (as determined by whether
Character.digit(char, int)returns a nonnegative value), except that the first character may be an ASCII minus sign'-' ('\u002D')to indicate a negative value or an ASCII plus sign'+' ('\u002B')to indicate a positive value. The resulting long value is returned.
将字符串参数解析为第二个参数指定的基数中的有符号长整型。字符串中的字符必须都是指定基数的数字（由是否
Character.digit(char, int)返回非负值决定），除了第一个字符可以是ASCII减号'-' ('\u002D')表示负值或ASCII加号'+' ('\u002B')表示正值.返回结果 long 值。
In order to get MAX_VALUEwe need:
为了得到MAX_VALUE我们需要：
String max = "1111111"; // 7 binary digits, +127
// or
String max2 = "+1111111"; // 7 binary digits, +127
回答by Evgeniy Dorofeev
This is because Long.parseLong cannot parse two's complement representation. The only way to parse two's complement binary string representation in Java SE is BigInteger:
这是因为 Long.parseLong 无法解析二进制补码表示。在 Java SE 中解析二进制补码二进制字符串表示的唯一方法是 BigInteger：
long l = new BigInteger("1000000000000000000000000000000000000000000000000000000000000000", 2).longValue()
this gives expected -9223372036854775808result
这给出了预期的 -9223372036854775808result
回答by SmRndGuy
Largest long value is actually:
最大的 long 值实际上是：
0111111111111111111111111111111111111111111111111111111111111111b = 9223372036854775807
回答by Enrico Giurin
This is the largest possible long (9223372036854775807 = 2 exp 63 - 1) in binary format. Note the L at the end of the last digit.
这是二进制格式的最大可能长度 (9223372036854775807 = 2 exp 63 - 1)。注意最后一位数字末尾的 L。
long largestLong = 0B0111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111L;
