需要对Java的32位整数表示系统进行说明吗？

Question

I need some clarifications regarding the binary representation of decimal in Java (or any other language for that matter).

pardon, if this is too basic, but I need to understand it thoroughly.

x = 31 is represented in Java 32 bit as:

x — > 00000000 00000000 00000000 00011111  // 

the binary to decimal conversion is achieved by:
                               2^4+2^3+2^2+2^1+2^0=31

Now, if you consider all the bits turned on, except the signed bit (most significant bit) , we get

y -> 01111111 11111111 11111111 11111111

and binary to decimal conversion by summing powers of 2 will be:
        2^31+2^30………+2^3+2^2+2^1+2^0=4294967295.

however, if you do:

System.out.println(Integer.parseInt("1111111111111111111111111111111",2));
you get: 2147483647 which is 2^31-1

so it means, the when 31 bits are turned on, I do not get the additive sum of the powers of 2. why is this?

may be I did not understand something, if some one could clarify that will be very helpful.

In case this helps: All the two raise powers till 31 are in this list:

[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648]

EDIT: I corrected the y -representation, now it has 32 bits, but if you calculate the 31 bits that are turned on with summing the powers, you would get 4294967295. I have one liner in python here

>>> reduce(lambda x,y: x+y, [pow(2,i) for i in range(32)])
4294967295

Answer 1

The problem is that here:

y -> 011111111 11111111 11111111 11111111

and binary to decimal conversion by summing powers of 2 will be:
        2^31+2^30………+2^3+2^2+2^1+2^0=4294967295.

you've put one too many 1s. As Jon Skeet mentioned in the comments, you should have only 31 1s, not 32. Thus, the sum should start at 2^30, not 2^31.

( Update : Well, you've updated that bit to have the correct number of 1s. My statement about the sum though, still stands. It should start at 2 ³⁰ , not 2 ³¹ . )

When you did this part:

System.out.println(Integer.parseInt("1111111111111111111111111111111",2));
you get: 2147483647 which is 2^31-1

you have the number of 1s correct (31).

Answer 2

so it means, the when 31 bits are turned on, I do not get the additive sum of the powers of 2.

Yes you do - you get 2 ⁰ + 2 ¹ + 2 ² + 2 ³ + ... + 2 ³⁰ ... which is 2 ³¹ - 1, exactly as you're seeing.

You're not adding 2 ³¹ because that would be represented by a 32nd 1 , at which point you'd be beyond the limit of what an int can represent in Java.

It's probably easier to consider smaller numbers. Suppose you have 5 bits - your example earlier on:

Integer.parseInt("11111")

That would print out 31, exactly as you said before... because 31 is 2 ⁵ - 1.

So for n bits all set to 1, you get 2 ⁿ - 1... which is correct for your example with 31 bits. There's nothing inconsistent here.

Answer 3

y -> 01111111 11111111 11111111 11111111
----------------------------------------
     31       23       15       7

The number is the index of the most significant bit in each byte.

So 0x2^31 + 1x2^30 + 1x2^29 + ... + 1x2^0 = 2147483647

You were simply going one bit too far.

You can actually test this

int a = 0b01111111_11111111_11111111_11111111;
System.out.println(a);

prints

2147483647