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junior
Asked: 2022-05-20 05:51:48 +0800 CST 2022-05-20 05:51:48 +0800 CST

Why is bit 32 negative?

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I am reviewing the bit offset in java but it seems very curious to me that bit 32 gives me a negative. I assume that it is due to the INT(32bits) data type and that it has a limit of bytes, but it would help me a lot if someone clarified it for me. .

For example:

here it moved 30 times, that is, it stayed in bit 31

System.out.println(0B1 << 30);// es 1073741824 en base 10

here I scroll 31 times I stay in bit 32

System.out.println(0B1 << 31);// es -2147483648 en base 10

here it moved 32 times, it stayed in bit 33, that is, it went to byte 5

System.out.println(0B1 << 32);// es 1 en base 10
java

1 Answers

  1. Best Answer

    Actually the last bit does not represent the sign, it is an error to make that approach. If it were that way, then the 1decimal value, using 4 bits to simplify instead of 32, would render 0001band the -1would render 1001b, and it doesn't. The value 1001brepresents the -7decimal value.

    What happens is that in 2's complement negative values ​​are represented by taking the positive value, negating all bits, and then adding 1. For example:

    The value 3, expressed in 4-bit, binary, is 0011b. To represent the value -3, all the bits are inverted, leaving then 1100band then adding 1b, giving the final result that is 1101b.

    As a side effect, all negative numbers have their most significant bit in value 1, and the most significant bit of your 32-bit integer in Java is bit 31.

    The largest negative number that can be represented in 2's complement is the one with its most significant bit in 1and the remainder in 0.

    The formal representation for the negative number of a 2's complement value Nis ( 2 ^n)-N , where n is the number of bits and N is the value to represent.

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