I am trying to understand how Java stores integer internally. I know all java primitive integers are signed, (except short?). That means one less bit available in a byte for the number.
My question is, are all integers (positive and negative) stored as two's complement or are only negative numbers in two's complement?
I see that the specs says x bit two's complement number. But I often get confused.
For instance:
int x = 15; // Stored as binary as is? 00000000 00000000 00000000 00001111?
int y = -22; // Stored as two complemented value? 11111111 11111111 11111111 11101010
Edit
To be clear, x = 15
In binary as is: `00000000 00000000 00000000 00001111'
Two's complement: `11111111 11111111 11111111 11110001`
So if your answer is all numbers are stored as two's complement then:
int x = 15; // 11111111 11111111 11111111 11110001
int y = -22 // 11111111 11111111 11111111 11101010
The confusion here again is the sign says, both are negative numbers. May be I am misreading / misunderstanding it?
Edit
Not sure my question is confusing. Forced to isolate the question:
My question precisely: Are positive numbers stored in binary as is while negative numbers are stored as two's complement?
Some said all are stored in two's complement and one answer says only negative numbers are stored as two's complement.
Let's start by summarizing Java primitive data types:
byte: Byte data type is an 8-bit signed two's complement integer.
Short: Short data type is a 16-bit signed two's complement integer.
int: Int data type is a 32-bit signed two's complement integer.
long: Long data type is a 64-bit signed two's complement integer.
float: Float data type is a single-precision 32-bit IEEE 754 floating point.
double: double data type is a double-precision 64-bit IEEE 754 floating point.
boolean: boolean data type represents one bit of information.
char: char data type is a single 16-bit Unicode character.
Source
Two's complement
"The good example is from wiki that the relationship to two's complement is realized by noting that 256 = 255 + 1, and (255 − x) is the ones' complement of x
0000 0111=7 two's complement is 1111 1001= -7
the way it works is the MSB(the most significant bit) receives a negative value so in the case above
-7 = 1001= -8 + 0+ 0+ 1
Positive integers are generally stored as simple binary numbers (1 is 1, 10 is 2, 11 is 3, and so on).
Negative integers are stored as the two's complement of their absolute value. The two's complement of a positive number is when using this notation a negative number.
Source
Since I received a few points for this answer, I decided to add more information to it.
A more detailed answer:
Among others there are four main approaches to represent positive and negative numbers in binary, namely:
Signed Magnitude
One's Complement
Two's Complement
Bias
1. Signed Magnitude
Uses the most significant bit to represent the sign, the remaining bits are used to represent the absolute value. Where 0 represents a positive number and 1 represents a negative number, example:
1011 = -3
0011 = +3
This representation is simpler. However, you cannot add binary numbers in the same way that you add decimal numbers, making it harder to be implemented at the hardware level. Moreover, this approach uses two binary patterns to represent the 0, -0 (1000) and +0 (0000).
2. One's Complement
In this representation, we invert all the bits of a given number to find out its complementary. For example:
010 = 2, so -2 = 101 (inverting all bits).
The problem with this representation is that there still exist two bits patterns to represent the 0, negative 0 (1111) and positive 0 (0000)
3. Two's Complement
To find the negative of a number, in this representation, we invert all the bits and then add one bit. Adding one bit solves the problem of having two bits patterns representing 0. In this representation, we only have one pattern for
0 (0000).
For example, we want to find the binary negative representation of 4 (decimal) using 4 bits. First, we convert 4 to binary:
4 = 0100
then we invert all the bits
0100 -> 1011
finally, we add one bit
1011 + 1 = 1100.
So 1100 is equivalent to -4 in decimal if we are using a Two's Complement binary representation with 4 bits.
A faster way to find the complementary is by fixing the first bit that as value 1 and inverting the remaining bits. In the above example it would be something like:
0100 -> 1100
^^
||-(fixing this value)
|--(inverting this one)
Two's Complement representation, besides having only one representation for 0, it also adds two binary values in the same way that in decimal, even numbers with different signs. Nevertheless, it is necessary to check for overflow cases.
4. Bias
This representation is used to represent the exponent in the IEEE 754 norm for floating points. It has the advantage that the binary value with all bits to zero represents the smallest value. And the binary value with all bits to 1 represents the biggest value. As the name indicates, the value is encoded (positive or negative) in binary with n bits with a bias (normally 2^(n-1) or 2^(n-1)-1).
So if we are using 8 bits, the value 1 in decimal is represented in binary using a bias of 2^(n-1), by the value:
+1 + bias = +1 + 2^(8-1) = 1 + 128 = 129
converting to binary
1000 0001
Java integers are of 32 bits, and always signed. This means, the most significant bit (MSB) works as the sign bit. The integer represented by an int is nothing but the weighted sum of the bits. The weights are assigned as follows:
Bit# Weight
31 -2^31
30 2^30
29 2^29
... ...
2 2^2
1 2^1
0 2^0
Note that the weight of the MSB is negative (the largest possible negative actually), so when this bit is on, the whole number (the weighted sum) becomes negative.
Let's simulate it with 4-bit numbers:
Binary Weighted sum Integer value
0000 0 + 0 + 0 + 0 0
0001 0 + 0 + 0 + 2^0 1
0010 0 + 0 + 2^1 + 0 2
0011 0 + 0 + 2^1 + 2^0 3
0100 0 + 2^2 + 0 + 0 4
0101 0 + 2^2 + 0 + 2^0 5
0110 0 + 2^2 + 2^1 + 0 6
0111 0 + 2^2 + 2^1 + 2^0 7 -> the most positive value
1000 -2^3 + 0 + 0 + 0 -8 -> the most negative value
1001 -2^3 + 0 + 0 + 2^0 -7
1010 -2^3 + 0 + 2^1 + 0 -6
1011 -2^3 + 0 + 2^1 + 2^0 -5
1100 -2^3 + 2^2 + 0 + 0 -4
1101 -2^3 + 2^2 + 0 + 2^0 -3
1110 -2^3 + 2^2 + 2^1 + 0 -2
1111 -2^3 + 2^2 + 2^1 + 2^0 -1
So, the two's complement thing is not an exclusive scheme for representing negative integers, rather we can say that the binary representation of integers are always the same, we just negate the weight of the most significant bit. And that bit determines the sign of the integer.
In C, there is a keyword unsigned (not available in java), which can be used for declaring unsigned int x;. In the unsigned integers, the weight of the MSB is positive (2^31) rather than being negative. In that case the range of an unsigned int is 0 to 2^32 - 1, while an int has range -2^31 to 2^31 - 1.
From another point of view, if you consider the two's complement of x as ~x + 1 (NOT x plus one), here's the explanation:
For any x, ~x is just the bitwise inverse of x, so wherever x has a 1-bit, ~x will have a 0-bit there (and vice versa). So, if you add these up, there will be no carry in the addition and the sum will be just an integer every bit of which is 1.
For 32-bit integers:
x + ~x = 1111 1111 1111 1111 1111 1111 1111 1111
x + ~x + 1 = 1111 1111 1111 1111 1111 1111 1111 1111 + 1
= 1 0000 0000 0000 0000 0000 0000 0000 0000
The leftmost 1-bit will simply be discarded, because it doesn't fit in 32-bits (integer overflow). So,
x + ~x + 1 = 0
-x = ~x + 1
So you can see that the negative x can be represented by ~x + 1, which we call the two's complement of x.
I have ran the following program to know it
public class Negative {
public static void main(String[] args) {
int i =10;
int j = -10;
System.out.println(Integer.toBinaryString(i));
System.out.println(Integer.toBinaryString(j));
}
}
Output is
1010
11111111111111111111111111110110
From the output it seems that it has been using two's complement.
Oracle provides some documentation regarding Java Datatypes that you may find interesting. Specifically:
int: The int data type is a 32-bit signed two's complement integer. It has a minimum value of -2,147,483,648 and a maximum value of 2,147,483,647 (inclusive).
Btw, short is also stored as two's complement.
Positive numbers are stored/retrived as it is.
e.g) For +ve number 10; byte representation will be like 0-000 0010
(0 - MSB will represent that it is +ve).
So while retrieving based on MSB; it says it is +ve,
so the value will be taken as it is.
But negative numbers will be stored after 2's complement (other than
MSB bit), and MSB bit will be set to 1.
e.g) when storing -10 then
0-000 0010 -> (1's complement) -> 0-111 1101
-> (2's complement) 0-111 1101 + 1 -> 0-111 1110
Now MSB will be set to one, since it is negative no -> 1-111 1110
when retrieving, it found that MSB is set to 1. So it is negative no.
And 2's complement will be performed other than MSB.
1-111 1110 --> 1-000 0001 + 1 --> 1-000 0010
Since MSB representing this is negative 10 --> hence -10 will be retrived.
Casting
Also note that when you are casting int/short to byte, only last byte will be considered along with last byte MSB,
Take example "-130" short, it might be stored like below
(MSB)1-(2's complement of)130(1000 0010) --> 1-111 1111 0111 1110
Now byte casting took last byte which is 0111 1110. (0-MSB)
Since MSB says it is +ve value, so it will be taken as it is.
Which is 126. (+ve).
Take another example "130" short, it might be stored like below
0-000 000 1000 0010 (MSB = 0)
Now byte casting took last byte which is 1000 0010 . (1=MSB)
Since MSB says it is -ve value, 2's complement will be performed and negative number will be returned. So in this case -126 will be returned.
1-000 0010 -> (1's complement) -> 1-111 1101
-> (2's complement) 1-111 1101 + 1 -> 1-111 1110 -> (-)111 1110
= -126
Diff between (int)(char)(byte) -1 AND (int)(short)(byte) -1
(byte)-1 -> 0-000 0001 (2's Comp) -> 0-111 1111 (add sign) -> 1-111 1111
(char)(byte)-1 -> 1-111 1111 1111 1111 (sign bit is carry forwarded on left)
similarly
(short)(byte)-1-> 1-111 1111 1111 1111 (sign bit is carry forwarded on left)
But
(int)(char)(byte)-1 -> 0-0000000 00000000 11111111 11111111 = 65535
since char is unsigned; MSB won't be carry forwarded.
AND
(int)(Short)(byte)-1 -> 1-1111111 11111111 11111111 11111111 = -1
since short is signed; MSB is be carry forwarded.
References
Why is two's complement used to represent negative numbers?
What is “2's Complement”?
The most significant bit (32nd) indicates that the number is positive or negative. If it is 0, it means the number is positive and it is stored in its actual binary representation. but if it is 1, it means the number is negative and is stored in its two's complement representation. So when we give weight -2^32 to the 32nd bit while restoring the integer value from its binary representation, We get the actual answer.
According to this document, all integers are signed and stored in two's complement format for java. Not certain of its reliability..
positive numbers are stored directly as binary. 2's compliment is required for negative numbers.
for example:
15 : 00000000 00000000 00000000 00001111
-15: 11111111 11111111 11111111 11110001
here is the difference in signed bit.
Thank you, dreamcrash for the answer https://stackoverflow.com/a/13422442/1065835; on the wiki page they give an example which helped me understand how to find out the binary representation of the negative counterpart of a positive number.
For example, using 1 byte (= 2 nibbles = 8 bits), the decimal number 5
is represented by
0000 01012 The most significant bit is 0, so the pattern represents a
non-negative value. To convert to −5 in two's-complement notation, the
bits are inverted; 0 becomes 1, and 1 becomes 0:
1111 1010 At this point, the numeral is the ones' complement of the
decimal value −5. To obtain the two's complement, 1 is added to the
result, giving:
1111 1011 The result is a signed binary number representing the
decimal value −5 in two's-complement form. The most significant bit is
1, so the value represented is negative.
For positive integer 2'complement value is same with MSB bit 0 (like +14 2'complement is 01110).
For only negative integer only we are calculating 2'complement value (-14= 10001+1 = 10010).
So final answer is both the values(+ve and -ve) are stored in 2'complement form only.
Related
Integer variables are 4-bytes or 32-bits, and 2^31 and -2^31 both in binary numbers are 32 bits. But when you put 2^31 = 2,147,483,648 in an integer variable it shows an error, but for -2^31 it is ok. Why?
Integer variables are 4-bytes or 32-bits, and 2^31 and -2^31 both in binary numbers are 32 bits
No they are not.
in basic binary, negative numbers aren't a thing. We have zeroes and ones. There is no - sign.
In binary, 2^31 becomes:
1000 0000 0000 0000 0000 0000 0000 0000
In binary, -2^31 cannot be represented without first defining how negative numbers are to be stored.
Commonly (and java does this too), a system called 2's complement is used. 2's complement sounds real complicated: Take the number, say, 5. Represent it in binary (for this exercise, let's go with byte, i.e. 8 bits): 0000 0101. Now, flip all bits: 1111 1010, and then add 1: 1111 1011.
That is -5 in signed 2's complement binary.
This bizarre system has two amazing properties: Math continues to work as normal without needing to know if the number is signed or unsigned. Let's try it. -5 + 2 is -3, right? let's see.. what's 1111 1011 + 0000 0010? Without worrying about 2's complement at all, I get 1111 1101. Let's apply 2's complement conversion: first flip the bits: 0000 0010, then add 1: 0000 0011, which is... 3. So -5 + 2 is -3. Check. The other amazing property is that it doesn't 'waste' 2 of the 2^32 "slots" on zeroes. Let's try the 2's complement of 0: 0000 0000, then flip all bits: 1111 1111, then add 1: 0000 0000 (with a bit overflow that we ignore). That's nice: 0 is its own 2's complement. We can't tell 0 and -0 apart, but that's generally a good thing.
Another property of this system is that the first bit is the 'sign' bit. if it is 1, it is negative, if 0, it is not.
Let's try to 2's complement 1000 0000 0000 0000 0000 0000 0000 0000. First, flip the bits: 0111 1111 1111 1111 1111 1111 1111 1111. Then add 1: 1000 0000 0000 0000 0000 0000 0000 0000. Wait. That's... what we had!!
Yup. and because the first bit is negative, 1000 0000 0000 0000 0000 0000 0000 0000 is NEGATIVE.
Perhaps you are forgetting that 0 is a thing, and 0 is neither positive nor negative.
So, if 0 needs to be representable, and gets a 0 sign bit (zero in bits is 0000000... of course), that means the 'space' in the half of all representable numbers that start with a 0 is now one smaller, because 0 has eaten one slot. That means there is one more negative number representable vs. the positive numbers. (or, alternatively, that 0 'counts' as positive, therefore 0 is the first positive number, but -1 is the first negative number). Therefore, there must be at least 1 negative number that has no positive equivalent in 2's complement. That number is... 2^31. -2^31 fits in 32-bit signed. +2^31 doesn't.
Let's imagine a 3-bit signed number, with 2's complement. We can list them all:
000 = 0
001 = 1
010 = 2
011 = 3
100 = -4
101 = -3
110 = -2
111 = -1
Note how -4 is in there, but +4 is not, and note how we covered 8 numbers. 2^3 = 8 - 3 bits can represent 8 numbers, not more than that.
From the oracle documentation we got that:
int: By default, the int data type is a 32-bit signed two's complement integer, which has a minimum value of -2^31 and a maximum value of 2^31-1. In Java SE 8 and later, you can use the int data type to represent an unsigned 32-bit integer, which has a minimum value of 0 and a maximum value of 2^32-1. Use the Integer class to use int data type as an unsigned integer. See the section The Number Classes for more information. Static methods like compareUnsigned, divideUnsigned etc have been added to the Integer class to support the arithmetic operations for unsigned integers.
From another document(simple and quite understandable) we got:
When an integer is signed, one of its bits becomes the sign bit, meaning that the maximum magnitude of the number is halved. (So an unsigned 32-bit int can store up to 2^32-1, whereas its signed counterpart has a maximum positive value of 2^31-1.)
In Java, all integer types are signed (except char).
Is because the first bit indicate the sign bit. Maximum positive value it can store it 2^31 - 1. There are many resources available for this.
Pretty basic stuff i am sure but bits are not my forte.
So for some internal calculation i am trying to convert a given input ( constraint is that it would be a integer string for sure) into its hex equivalent, what stumped me is on how to get
Hex signed 2's complement:
My noob code:
private String toHex(String arg, boolean isAllInt) {
String hexVal = null;
log.info("arg {}, isAllInt {}", arg, isAllInt);
if (isAllInt) {
int intVal = Integer.parseInt(arg);
hexVal = Integer.toHexString(intVal);
// some magic to convert this hexVal to its 2's compliment
} else {
hexVal = String.format("%040x", new BigInteger(1, arg.getBytes(StandardCharsets.UTF_8)));
}
log.info("str {} hex {}", arg, hexVal);
return hexVal;
}
Input: 00001
Output: 1
Expected Output: 0001
Input: 00216
Output: D8
Expected Output: 00D8
00216
Input: 1192633166
Output: 4716234E
Expected Output: 4716234E
any predefined library is much welcome or any other useful pointers!
So to pad the hex digits up to either 4 digits or 8 digits, do:
int intVal = Integer.parseInt(arg);
if (intVal >= 0 && intVal <= 0xffff) {
hexVal = String.format("%04x", intVal);
} else {
hexVal = String.format("%08x", intVal);
}
See Java documentation on how the format strings work.
Answering the two's complement aspect.
Two's Complement Representation
Two's complement is an agreement how to represent signed integral numbers in e.g. 16 bits (in olden times, different representations have been used by various processors, e.g. one's complement or sign-magnitude).
Positive numbers and zero are represented as expected:
0 is 0000 0000 0000 0000 or hex 0000
1 is 0000 0000 0000 0001 or hex 0001
2 is 0000 0000 0000 0010 or hex 0002
3 is 0000 0000 0000 0011 or hex 0003
4 is 0000 0000 0000 0100 or hex 0004
Negative numbers are represented by adding 1 0000 0000 0000 0000 to them, giving:
-1 is 1111 1111 1111 1111 or hex ffff
-2 is 1111 1111 1111 1110 or hex fffe
-3 is 1111 1111 1111 1101 or hex fffd
This is equivalent to: take the positive representation, flip all bits, and add 1.
For negative numbers, the highest bit is always 1. And that's how the machine distinguishes positive and negative numbers.
All processors in use today do their integer arithmetic based on two's complement representation, so there's typically no need to do special tricks. All the Java datatypes like byte, short, int, and long are defined to be signed numbers in two's complement representation.
In a comment you wrote
2's compliment is hex of negative of original value
That mixes up the concepts a bit. Two's complement is basically defined on bit patterns, and groups of 4 bits from these bit patterns can nicely be written as hex digits. Two's complement is about representing negative values as bit patterns, but from your question and comments I read that you don't expect negative values, so two's complement shouldn't concern you.
Hex Strings
To represent signed values as hex strings, Java (and most other languages / environments) simply looks at the bit patterns, ignoring their positive / negative interpretation, meaning that e.g. -30 (1111 1111 1110 0010) does not get shown as "-1e" with a minus sign, but as "ffe2".
Because of this, negative values will always get translated to a string with maximum length according to the value's size (16 bits, 32 bits, 64 bits giving 4, 8, or 16 hex digits), because the highest bit will be 1, resulting in a leading hex digit surely not being zero. So for negative values, there's no need to do any padding.
Small positive values will have leading zeros in their hex representation, and Java's toHexString() method suppresses them, so 1 (0000 0000 0000 0001) becomes "1" and not "0001". That's why e.g. format("%04x", ...), as in #nos's answer, is useful.
I'm a little confused by the ~ operator. Code goes below:
a = 1
~a #-2
b = 15
~b #-16
How does ~ do work?
I thought, ~a would be something like:
0001 = a
1110 = ~a
why not?
You are exactly right. It's an artifact of two's complement integer representation.
In 16 bits, 1 is represented as 0000 0000 0000 0001. Inverted, you get 1111 1111 1111 1110, which is -2. Similarly, 15 is 0000 0000 0000 1111. Inverted, you get 1111 1111 1111 0000, which is -16.
In general, ~n = -n - 1
The '~' operator is defined as:
"The bit-wise inversion of x is defined as -(x+1). It only applies to integral numbers."Python Doc - 5.5
The important part of this sentence is that this is related to 'integral numbers' (also called integers). Your example represents a 4 bit number.
'0001' = 1
The integer range of a 4 bit number is '-8..0..7'. On the other hand you could use 'unsigned integers', that do not include negative number and the range for your 4 bit number would be '0..15'.
Since Python operates on integers the behavior you described is expected. Integers are represented using two's complement. In case of a 4 bit number this looks like the following.
7 = '0111'
0 = '0000'
-1 = '1111'
-8 = '1000'
Python uses 32bit for integer representation in case you have a 32-bit OS. You can check the largest integer with:
sys.maxint # (2^31)-1 for my system
In case you would like an unsigned integer returned for you 4 bit number you have to mask.
'0001' = a # unsigned '1' / integer '1'
'1110' = ~a # unsigned '14' / integer -2
(~a & 0xF) # returns 14
If you want to get an unsigned 8 bit number range (0..255) instead just use:
(~a & 0xFF) # returns 254
It looks like I found simpler solution that does what is desired:
uint8: x ^ 0xFF
uint16: x ^ 0xFFFF
uint32: x ^ 0xFFFFFFFF
uint64: x ^ 0xFFFFFFFFFFFFFFFF
You could also use unsigned ints (for example from the numpy package) to achieve the expected behaviour.
>>> import numpy as np
>>> bin( ~ np.uint8(1))
'0b11111110'
The problem is that the number represented by the result of applying ~ is not well defined as it depends on the number of bits used to represent the original value. For instance:
5 = 101
~5 = 010 = 2
5 = 0101
~5 = 1010 = 10
5 = 00101
~5 = 11010 = 26
However, the two's complement of ~5 is the same in all cases:
two_complement(~101) = 2^3 - 2 = 6
two_complement(~0101) = 2^4 - 10 = 6
two_complement(~00101) = 2^5 - 26 = 6
And given that the two's complement is used to represent negative values, it makes sense to consider ~5 as the negative value, -6, of its complement.
So, more formally, to arrive at this result we have:
flipped zeros and ones (that's equivalent to taking the ones' complement)
taken two's complement
applied negative sign
and if x is a n-digit number:
~x = - two_complement(one_complement(x)) = - two_complement(2^n - 1 - x) = - (2^n - (2^n - 1 - x)) = - (x + 1)
I am trying to understand how Java stores integer internally. I know all java primitive integers are signed, (except short?). That means one less bit available in a byte for the number.
My question is, are all integers (positive and negative) stored as two's complement or are only negative numbers in two's complement?
I see that the specs says x bit two's complement number. But I often get confused.
For instance:
int x = 15; // Stored as binary as is? 00000000 00000000 00000000 00001111?
int y = -22; // Stored as two complemented value? 11111111 11111111 11111111 11101010
Edit
To be clear, x = 15
In binary as is: `00000000 00000000 00000000 00001111'
Two's complement: `11111111 11111111 11111111 11110001`
So if your answer is all numbers are stored as two's complement then:
int x = 15; // 11111111 11111111 11111111 11110001
int y = -22 // 11111111 11111111 11111111 11101010
The confusion here again is the sign says, both are negative numbers. May be I am misreading / misunderstanding it?
Edit
Not sure my question is confusing. Forced to isolate the question:
My question precisely: Are positive numbers stored in binary as is while negative numbers are stored as two's complement?
Some said all are stored in two's complement and one answer says only negative numbers are stored as two's complement.
Let's start by summarizing Java primitive data types:
byte: Byte data type is an 8-bit signed two's complement integer.
Short: Short data type is a 16-bit signed two's complement integer.
int: Int data type is a 32-bit signed two's complement integer.
long: Long data type is a 64-bit signed two's complement integer.
float: Float data type is a single-precision 32-bit IEEE 754 floating point.
double: double data type is a double-precision 64-bit IEEE 754 floating point.
boolean: boolean data type represents one bit of information.
char: char data type is a single 16-bit Unicode character.
Source
Two's complement
"The good example is from wiki that the relationship to two's complement is realized by noting that 256 = 255 + 1, and (255 − x) is the ones' complement of x
0000 0111=7 two's complement is 1111 1001= -7
the way it works is the MSB(the most significant bit) receives a negative value so in the case above
-7 = 1001= -8 + 0+ 0+ 1
Positive integers are generally stored as simple binary numbers (1 is 1, 10 is 2, 11 is 3, and so on).
Negative integers are stored as the two's complement of their absolute value. The two's complement of a positive number is when using this notation a negative number.
Source
Since I received a few points for this answer, I decided to add more information to it.
A more detailed answer:
Among others there are four main approaches to represent positive and negative numbers in binary, namely:
Signed Magnitude
One's Complement
Two's Complement
Bias
1. Signed Magnitude
Uses the most significant bit to represent the sign, the remaining bits are used to represent the absolute value. Where 0 represents a positive number and 1 represents a negative number, example:
1011 = -3
0011 = +3
This representation is simpler. However, you cannot add binary numbers in the same way that you add decimal numbers, making it harder to be implemented at the hardware level. Moreover, this approach uses two binary patterns to represent the 0, -0 (1000) and +0 (0000).
2. One's Complement
In this representation, we invert all the bits of a given number to find out its complementary. For example:
010 = 2, so -2 = 101 (inverting all bits).
The problem with this representation is that there still exist two bits patterns to represent the 0, negative 0 (1111) and positive 0 (0000)
3. Two's Complement
To find the negative of a number, in this representation, we invert all the bits and then add one bit. Adding one bit solves the problem of having two bits patterns representing 0. In this representation, we only have one pattern for
0 (0000).
For example, we want to find the binary negative representation of 4 (decimal) using 4 bits. First, we convert 4 to binary:
4 = 0100
then we invert all the bits
0100 -> 1011
finally, we add one bit
1011 + 1 = 1100.
So 1100 is equivalent to -4 in decimal if we are using a Two's Complement binary representation with 4 bits.
A faster way to find the complementary is by fixing the first bit that as value 1 and inverting the remaining bits. In the above example it would be something like:
0100 -> 1100
^^
||-(fixing this value)
|--(inverting this one)
Two's Complement representation, besides having only one representation for 0, it also adds two binary values in the same way that in decimal, even numbers with different signs. Nevertheless, it is necessary to check for overflow cases.
4. Bias
This representation is used to represent the exponent in the IEEE 754 norm for floating points. It has the advantage that the binary value with all bits to zero represents the smallest value. And the binary value with all bits to 1 represents the biggest value. As the name indicates, the value is encoded (positive or negative) in binary with n bits with a bias (normally 2^(n-1) or 2^(n-1)-1).
So if we are using 8 bits, the value 1 in decimal is represented in binary using a bias of 2^(n-1), by the value:
+1 + bias = +1 + 2^(8-1) = 1 + 128 = 129
converting to binary
1000 0001
Java integers are of 32 bits, and always signed. This means, the most significant bit (MSB) works as the sign bit. The integer represented by an int is nothing but the weighted sum of the bits. The weights are assigned as follows:
Bit# Weight
31 -2^31
30 2^30
29 2^29
... ...
2 2^2
1 2^1
0 2^0
Note that the weight of the MSB is negative (the largest possible negative actually), so when this bit is on, the whole number (the weighted sum) becomes negative.
Let's simulate it with 4-bit numbers:
Binary Weighted sum Integer value
0000 0 + 0 + 0 + 0 0
0001 0 + 0 + 0 + 2^0 1
0010 0 + 0 + 2^1 + 0 2
0011 0 + 0 + 2^1 + 2^0 3
0100 0 + 2^2 + 0 + 0 4
0101 0 + 2^2 + 0 + 2^0 5
0110 0 + 2^2 + 2^1 + 0 6
0111 0 + 2^2 + 2^1 + 2^0 7 -> the most positive value
1000 -2^3 + 0 + 0 + 0 -8 -> the most negative value
1001 -2^3 + 0 + 0 + 2^0 -7
1010 -2^3 + 0 + 2^1 + 0 -6
1011 -2^3 + 0 + 2^1 + 2^0 -5
1100 -2^3 + 2^2 + 0 + 0 -4
1101 -2^3 + 2^2 + 0 + 2^0 -3
1110 -2^3 + 2^2 + 2^1 + 0 -2
1111 -2^3 + 2^2 + 2^1 + 2^0 -1
So, the two's complement thing is not an exclusive scheme for representing negative integers, rather we can say that the binary representation of integers are always the same, we just negate the weight of the most significant bit. And that bit determines the sign of the integer.
In C, there is a keyword unsigned (not available in java), which can be used for declaring unsigned int x;. In the unsigned integers, the weight of the MSB is positive (2^31) rather than being negative. In that case the range of an unsigned int is 0 to 2^32 - 1, while an int has range -2^31 to 2^31 - 1.
From another point of view, if you consider the two's complement of x as ~x + 1 (NOT x plus one), here's the explanation:
For any x, ~x is just the bitwise inverse of x, so wherever x has a 1-bit, ~x will have a 0-bit there (and vice versa). So, if you add these up, there will be no carry in the addition and the sum will be just an integer every bit of which is 1.
For 32-bit integers:
x + ~x = 1111 1111 1111 1111 1111 1111 1111 1111
x + ~x + 1 = 1111 1111 1111 1111 1111 1111 1111 1111 + 1
= 1 0000 0000 0000 0000 0000 0000 0000 0000
The leftmost 1-bit will simply be discarded, because it doesn't fit in 32-bits (integer overflow). So,
x + ~x + 1 = 0
-x = ~x + 1
So you can see that the negative x can be represented by ~x + 1, which we call the two's complement of x.
I have ran the following program to know it
public class Negative {
public static void main(String[] args) {
int i =10;
int j = -10;
System.out.println(Integer.toBinaryString(i));
System.out.println(Integer.toBinaryString(j));
}
}
Output is
1010
11111111111111111111111111110110
From the output it seems that it has been using two's complement.
Oracle provides some documentation regarding Java Datatypes that you may find interesting. Specifically:
int: The int data type is a 32-bit signed two's complement integer. It has a minimum value of -2,147,483,648 and a maximum value of 2,147,483,647 (inclusive).
Btw, short is also stored as two's complement.
Positive numbers are stored/retrived as it is.
e.g) For +ve number 10; byte representation will be like 0-000 0010
(0 - MSB will represent that it is +ve).
So while retrieving based on MSB; it says it is +ve,
so the value will be taken as it is.
But negative numbers will be stored after 2's complement (other than
MSB bit), and MSB bit will be set to 1.
e.g) when storing -10 then
0-000 0010 -> (1's complement) -> 0-111 1101
-> (2's complement) 0-111 1101 + 1 -> 0-111 1110
Now MSB will be set to one, since it is negative no -> 1-111 1110
when retrieving, it found that MSB is set to 1. So it is negative no.
And 2's complement will be performed other than MSB.
1-111 1110 --> 1-000 0001 + 1 --> 1-000 0010
Since MSB representing this is negative 10 --> hence -10 will be retrived.
Casting
Also note that when you are casting int/short to byte, only last byte will be considered along with last byte MSB,
Take example "-130" short, it might be stored like below
(MSB)1-(2's complement of)130(1000 0010) --> 1-111 1111 0111 1110
Now byte casting took last byte which is 0111 1110. (0-MSB)
Since MSB says it is +ve value, so it will be taken as it is.
Which is 126. (+ve).
Take another example "130" short, it might be stored like below
0-000 000 1000 0010 (MSB = 0)
Now byte casting took last byte which is 1000 0010 . (1=MSB)
Since MSB says it is -ve value, 2's complement will be performed and negative number will be returned. So in this case -126 will be returned.
1-000 0010 -> (1's complement) -> 1-111 1101
-> (2's complement) 1-111 1101 + 1 -> 1-111 1110 -> (-)111 1110
= -126
Diff between (int)(char)(byte) -1 AND (int)(short)(byte) -1
(byte)-1 -> 0-000 0001 (2's Comp) -> 0-111 1111 (add sign) -> 1-111 1111
(char)(byte)-1 -> 1-111 1111 1111 1111 (sign bit is carry forwarded on left)
similarly
(short)(byte)-1-> 1-111 1111 1111 1111 (sign bit is carry forwarded on left)
But
(int)(char)(byte)-1 -> 0-0000000 00000000 11111111 11111111 = 65535
since char is unsigned; MSB won't be carry forwarded.
AND
(int)(Short)(byte)-1 -> 1-1111111 11111111 11111111 11111111 = -1
since short is signed; MSB is be carry forwarded.
References
Why is two's complement used to represent negative numbers?
What is “2's Complement”?
The most significant bit (32nd) indicates that the number is positive or negative. If it is 0, it means the number is positive and it is stored in its actual binary representation. but if it is 1, it means the number is negative and is stored in its two's complement representation. So when we give weight -2^32 to the 32nd bit while restoring the integer value from its binary representation, We get the actual answer.
According to this document, all integers are signed and stored in two's complement format for java. Not certain of its reliability..
positive numbers are stored directly as binary. 2's compliment is required for negative numbers.
for example:
15 : 00000000 00000000 00000000 00001111
-15: 11111111 11111111 11111111 11110001
here is the difference in signed bit.
Thank you, dreamcrash for the answer https://stackoverflow.com/a/13422442/1065835; on the wiki page they give an example which helped me understand how to find out the binary representation of the negative counterpart of a positive number.
For example, using 1 byte (= 2 nibbles = 8 bits), the decimal number 5
is represented by
0000 01012 The most significant bit is 0, so the pattern represents a
non-negative value. To convert to −5 in two's-complement notation, the
bits are inverted; 0 becomes 1, and 1 becomes 0:
1111 1010 At this point, the numeral is the ones' complement of the
decimal value −5. To obtain the two's complement, 1 is added to the
result, giving:
1111 1011 The result is a signed binary number representing the
decimal value −5 in two's-complement form. The most significant bit is
1, so the value represented is negative.
For positive integer 2'complement value is same with MSB bit 0 (like +14 2'complement is 01110).
For only negative integer only we are calculating 2'complement value (-14= 10001+1 = 10010).
So final answer is both the values(+ve and -ve) are stored in 2'complement form only.
The binary representation of Integer.MIN_VALUE in Java is 10000000000000000000000000000000.
Why wouldn't it be 10000000000000000000000000000001 instead, since the addition of one to its two's complement would be larger than 10000000000000000000000000000000 by one?
int in Java is a signed 32-bit number, meaning that the 32th bit, like you said, represents the negative number, in decimal, -(2^31), and in binary:
10000000000000000000000000000000
^ → "negative bit"
But the remaining 31 bits are still "positive" bits in a signed system, so:
10000000000000000000000000000001 → -2^31 + 1
And -2^31 + 1 is bigger than -2^31.