Are hexadecimal numbers ever negative? If yes then how?
For binary you would have signed and unsigned.
How would one represent them in Hex? I need this for a hex routine I am about to embark upon.
Yes. For example you'd have the following representations in signed 32-bit binary and hex:
Decimal: 1
Binary: 00000000 00000000 00000000 00000001
Hex: 00 00 00 01
Decimal: -1
Binary: 11111111 11111111 11111111 11111111
Hex: FF FF FF FF
Decimal: -2
Binary: 11111111 11111111 11111111 11111110
Hex: FF FF FF FE
As you can see, the Hex representation of negative numbers is directly related to the binary representation.
The high bit of a number determines if it is negative. So for instance an int is 32 bits long, so if bit 31 is a 1 it is negative. Now how you display that value be it hexadecimal or decimal doesn't matter. so the hex values like
0x80000000
0x91345232
0xA3432032
0xBFF32042
0xC33252DD
0xE772341F
0xFFFFFFFF
are all negative, because the top bit is set to 1
|
v
0x8 -> 1000
0x9 -> 1001
0xA -> 1010
0xB -> 1011
0xC -> 1100
0xD -> 1101
0xE -> 1110
0xF -> 1111
Yes they can be. It's the same as binary as to how you interpret it (signed vs unsigned).
You would take the binary signed and unsigned forms then represent them in hex as you would any binary number.
On one hand, why not - it's just a positional numeric system, like decimal.
On the other hand, they normally use hex notation to deduce the underlying bit pattern - and that is much more straightforward if the number is interpreted as unsigned.
So the answer is - it's possible, mathematically correct and internally consistent, but defeats the most common purpose of hex notation.
In Java, these are the bounds of the Integer data type:
Integer.MIN_VALUE = 0x80000000;
Integer.MAX_VALUE = 0x7fffffff;
Related
public void int2byte(){
int x = 128;
byte y = (byte) x;
System.out.println(Integer.toHexString(y));
}
I got result ffffff80, why not 80?
I got result 7f when x = 127.
Bytes are signed in Java. When you cast 128 to a byte, it becomes -128.
The int 128: 00000000 00000000 00000000 10000000
The byte -128: 10000000
Then, when you widen it back to an int with Integer.toHexString, because it's now negative, it gets sign-extended. This means that a bunch of 1 bits show up, explaining your extra f characters in the hex output.
You can convert it in an unsigned fashion using Byte.toUnsignedInt to prevent the sign extension.
Converts the argument to an int by an unsigned conversion. In an unsigned conversion to an int, the high-order 24 bits of the int are zero and the low-order 8 bits are equal to the bits of the byte argument.
System.out.println(Integer.toHexString(Byte.toUnsignedInt(y)));
An alternative is to bit-mask out the sign-extended bits by manually keeping only the 8 bits of the byte:
System.out.println(Integer.toHexString(y & 0xFF));
Either way, the output is:
80
This is because a positive value of 128 cannot be represented with a signed byte
The range is -128 to 127
So your int becomes sign-extended so instead of 0x00000080 you get 0xffffff80
Edit: As others have explained:
Integer 128 shares the last 8 bits with the byte (-128)
The difference is that Integer 128 has leading zeros.
0x00000080 (0000 0000 0000 0000 0000 0000 1000 0000)
while the byte -128 is just
0x80 (1000 0000)
When you convert byte -128 using Integer.toHexString() it takes the leading 1 and sign-extends it so you get
0xffffff80 (1111 1111 1111 1111 1111 1111 1000 0000)
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.
This is the code I've tried:
int num = ~0;
System.out.print(num);
Output: -1
From what I understand, ~ inverts the bits. So, 0000 0000 would become 1111 1111. How is this -1? I realize that this is a very basic question that involves two's complement, but I'm not able to figure it out.
Because -1 is represented as all ones.
System.out.println(Integer.toBinaryString(-1));
Output is
11111111111111111111111111111111
numbers are reprezented in 32 bit format.
To understand why it shows up as all 1 and then get converted to -1.
Reason:
~0 = ~(00000000 00000000 00000000 00000000) = (11111111 11111111 11111111 11111111) = -1
To understand more, please read this thread: How does the bitwise complement (~) operator work?
1111 1111 is in fact -1 and 1111 1110 is -2. Such is life, not sure how else to put it
If I have an byte value 00101011 and I want to represent it as int value I have: 00000000000000000000000000101011, what seems to be clear for me. Bu I have problem with byte value 11010100 which as int is represented by 11111111111111111111111111010100. I don't know where it came from. Can someone explain me the idea of extension byte value to int?
The byte value 11010100 represents a negative number (-44), because the most significant bit is set. When this undergoes primitive widening conversion, it must still represent the same negative value in the two's complement representation. This done using sign extension. That means that all new bits are the same as the original sign bit.
11010100 => -44
11111111 11111111 11111111 11010100 => -44
If this did not occur, then the sign bit would no longer be a sign bit, and it would be interpreted "normally", and the value would no longer be the same.
00000000 00000000 00000000 11010100 => 212
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.