在Java中将64位无符号int（以十六进制字符串形式）转换为十进制

Question

I have a hex string that represents a 64-bit unsigned int.

I want to convert this to a decimal representation in java.

I wanted to use Long.parseLong("HexString", 16); but in java the leftmost bit is the sign bit and any number greater than 2^31-1 would not fit.

SO how do i convert and store this data?

Thanks, Sunny

Answer 1

You can create a BigInteger using the constructor that takes a String and an int radix .

BigInteger bigInt = new BigInteger(hexString, 16);

If you have Java 8, you can use Long.parseUnsignedLong , which will allow you to parse all 64 bits, even if the bit that would normally put it over Long.MAX_VALUE is set. It still returns a long , which in Java is still signed, but it will be parsed as if it were an unsigned long .

long stillSigned = Long.parseUnsignedLong(hexString, 16);

If the value that gets parsed is negative, then the true unsigned value is the signed value + 2 ⁶⁴ .