i was thinking about size of int(signed and unsigned). The size of unsigned integer is very obvious to calculate but when it came to signed i got confused.. like if int is of 2 bytes and 15th byte is for sign then its size is 32767 thats understandable .. but when i tried to calculate the -ve minimum i did this 1 1111111111111..(16 ones). ignoring the signed bit(16th) the value was still 32767 so i got range +32767 to -32767(and thats not the right range) .
or maybe this is not the right way to calculate the range because compiler will take it as -1.
so how +32767 to -32768 is calculated ??
but when i tried to calculate the -ve minimum i did this 1 1111111111111..(16 ones). ignoring the signed bit(16th) the value was still 32767 so i got range +32767 to -32767(and thats not the right range)
Well, it can be the right range if your architecture defines the negative range the same way you do: that is, if it uses a "sign and magnitude" (or "sign and absolute value") representation.
That's completely valid: in C, signed integers are NOT necessarily represented using 2's complement (as you seem to assume).
If your platform, however, does indeed use 2's complement, then your method for calculating the minimal negative value is incorrect. Then you have to count backwards, from 1111 1111 1111 1111 (which is -1), until 1000 0000 0000 0000, which turns out to be -32768.
This is true because all of the 15 bits are changing (only the 0th bit, the sign bit stays the same as you can see), so there are 2 ^ 15 possible variations, and if you start from -1, you will arrive at -32768.
For twos complement representation, the negation of a number is achieved by taking the complement of the number and then adding 1. So, the number 0xFFFF complemented is 0x0000, and adding 1 yields 0x0001. So in twos complement, -1 is represented by the binary value of all bits set.
The largest positive signed 16 bit number is 0x7FFF, and its negative representation is 0x8001. Looking at that negative number, you can see that by subtracting one more from that value, we get negative number of one larger magnitude: 0x8000.
So, 32767 = 0x7FFF, and -32768 = 0x8000.
Related
I dont quite understand, if someone can provide examples to help me understand better. It would be greatly appreciated.
On a system using 2's complement and a 32-bit int, the range of values it can hold is -2147483648 to 2147483647.
If you were to negate the smallest possible int, i.e. -2147483648, the result would be 2147483648 which is out of range.
A sign-and-magnitude system cannot overflow in this way because 1 bit is reserved solely as a sign bit while the remaining bits (assuming no padding) are the value.
You can't get an overflow in sign-magnitude format. Negating the number simply inverts the sign and keeps the magnitude the same.
In two's complement, you get an overflow if you try to negate the most negative value, because there's always 1 more negative value than positive values. For instance, in 8 bits the range of values is -128 to 127. Since there's no 128, you get an overflow if you try to negate -128.
As 2's complement numbers do not have negative zero - they have only one zero so the number of the negative numbers is larger than the number of the positive numbers.
If you negate the minimum negative number you will get number which cannot be represented as a positive number of the same length.
My textbook provides the following explanation for the two's complement method for signed integers:
We’ll discuss this method as it applies to a 1-byte value. In that
context, the values 0 through 127 are represented by the last 7 bits,
with the high-order bit set to 0. So far, that’s the same as the
sign-magnitude method. Also, if the high-order bit is 1, the value is
negative. The difference comes in determining the value of that
negative number. Subtract the bit-pattern
for a negative number from the 9-bit pattern 100000000 (256 as
expressed in binary), and
the result is the magnitude of the value.
None of this makes any sense to me. Typical processors use octets (8-bit bytes). What does it mean by subtracting the 8-bit byte from the 9-bit byte?
Basically, for efficient computation, you want to have the same operations (addition, subtraction, etc.) to be performed the same way, regardless of the sign. So first, consider the case of unsigned bytes.
With 8 bits, you can represent any value between 0-255. Addition and subtraction work the same way as usual (modulo 256), and everything is fine.
Now, imagine that when you are at 127, incrementing by 1 gives you -128. We're still counting modulo 256, but the top 128 numbers have shifted by 256. Now, let's examine addition:
10 + (-5) = 10 + 251 (unsigned) = 261 = 5 (modulo 255).
Everything works as expected. So, in our new representation, -128 is 127 + 1, which is 01111111 + 1 which is 10000000. -1 will be 11111111. I hope I've helped.
You have 1 byte (8 digit number with each digit being a 0 or a 1)
2's complement works by looking at the first digit
10010011
^ this one
If it's a 0, then the number is positive and you can continue to convert binary to decimal normally.
If it's 1, then it's negative. Convert it normally (bin(10010011) = 147) and THEN subtract 256 (147 - 256 = -109), and there is your 2's complement number.
My memory about two's complement says:
"Flip the bits plus one"
Indeed, the following makes a number negative:
int i, j= 127;
i = (~j)+1;
printf ("%d\n",i); // prints -127
I read that the range of an int is dependent on a byte.
So taking int to be 4 bytes long, thats 4 * 8 bits = 32 bits.
So the range should be : 2 ^ (32-1) = 2 ^ (31)
Why do some people say its 2^31 - 1 though?
Thanks!
Because the counting starts from 0
And the range of int is 2,147,483,647 and 2^32 which is 2,147,483,648. hence we subtract 1
Also the loss of 1 bit is for the positive and negative sign
Check this interestinf wiki article on Integers:-
The most common representation of a positive integer is a string of
bits, using the binary numeral system. The order of the memory bytes
storing the bits varies; see endianness. The width or precision of an
integral type is the number of bits in its representation. An integral
type with n bits can encode 2n numbers; for example an unsigned type
typically represents the non-negative values 0 through 2n−1. Other
encodings of integer values to bit patterns are sometimes used, for
example Binary-coded decimal or Gray code, or as printed character
codes such as ASCII.
There are four well-known ways to represent signed numbers in a binary
computing system. The most common is two's complement, which allows a
signed integral type with n bits to represent numbers from −2(n−1)
through 2(n−1)−1. Two's complement arithmetic is convenient because
there is a perfect one-to-one correspondence between representations
and values (in particular, no separate +0 and −0), and because
addition, subtraction and multiplication do not need to distinguish
between signed and unsigned types. Other possibilities include offset
binary, sign-magnitude, and ones' complement.
You mean 232-1, NOT 232-1.
But your question is about why people use 231. The loss of a whole bit is if the int is a signed one. You lose the first bit to indicate if the number is positive or negative.
A signed int (32 bit) ranges from -2,147,483,648 to +2,147,483,647.
An unsigned int (32 bit) ranges from 0 to 4,294,967,295 (which is 232 -1).
int is a signed data type.
The first bit represents the sign, followed by bits for the value.
If the sign bit is 0, the value is simply the sum of all bits set to 1 ( to the power of 2).
e.g. 0...00101 is 20 + 22 = 5
if the first bit is 1, the value is -232 + the sum of all bits set to 1 (to the power of 2).
e.g. 1...111100 is -232 + 231 + 230 + ... + 22 = -4
all 0 will this result in zero.
When you calculate after, you will see that any number between (and including) the range - 231 and 20 + ... + 231 = 232 - 1 can be created with those 32 bits.
232-1 is not same as 232 - 1 (as 0 is included in the range, we subtract 1)
For your understanding, let us replace by small number 4 instead of 32
24-1 = 8
whereas 24-1 = 16-1 = 15.
Hope this helps!
Since integer is 32 bit. It could store total 2^32 values. So an integer ranges from -2^31 to 2^31-1 giving a total of 2^32 values(2^31 values in the negative range+2^31 values in positive range including 0).However, the first bit(the most significant bit) is reserved for the sign of the integer. Again u need to understand how negative integers are stored.They are stored in 2's complement form, So -9 will be stored as 2's complement of 9.
So 9 is stored in 32 bit system as
0000 0000 0000 0000 0000 0000 0000 1001
and -9 will be stored as
1111 1111 1111 1111 1111 1111 1111 0111 (2's complement of 9).
Again due to some arithmetic operation on an integer, if it happens to exceed the maximum value(2^31-1) then it will recycle to the negative values. So if you add 1 to 2^31-1 it will give you -2^31.
What does signed mean in C? I have this table to show:
This says signed char 128 to +127. 128 is also a positive integer, so how can this be something like +128 to +127? Or do 128 and +127 have different meanings? I am referring to the book Apress Beginning C.
A signed integer can represent negative numbers; unsigned cannot.
Signed integers have undefined behavior if they overflow, while unsigned integers wrap around using modulo.
Note that that table is incorrect. First off, it's missing the - signs (such as -128 to +127). Second, the standard does not guarantee that those types must fall within those ranges.
By default, numerical values in C are signed, which means they can be both negative and positive. Unsigned values on the other hand, don't allow negative numbers.
Because it's all just about memory, in the end all the numerical values are stored in binary. A 32 bit unsigned integer can contain values from all binary 0s to all binary 1s. When it comes to 32 bit signed integer, it means one of its bits (most significant) is a flag, which marks the value to be positive or negative. So, it's the interpretation issue, which tells that value is signed.
Positive signed values are stored the same way as unsigned values, but negative numbers are stored using two's complement method.
If you want to write negative value in binary, first write positive number, next invert all the bits and last add 1. When a negative value in two's complement is added to a positive number of the same magnitude, the result will be 0.
In the example below lets deal with 8-bit numbers, because it'll be simple to inspect:
positive 95: 01011111
negative 95: 10100000 + 1 = 10100001 [positive 161]
0: 01011111 + 10100001 = 100000000
^
|_______ as we're dealing with 8bit numbers,
the 8 bits which means results in 0
The table is missing the minuses. The range of signed char is -128 to +127; likewise for the other types on the table.
It was a typo in the book; signed char goes from -128 to 127.
Signed integers are stored using the two's complement representation, in which the first bit is used to indicate the sign.
In C, chars are just 8 bit integers. This means that they can go from -(2^7) to 2^7 - 1. That's because we use the 7 last bits for the number and the first bit for the sign. 0 means positive and 1 means negative (in two's complement representation).
The biggest positive 7 bit number is (01111111)b = 2^7 - 1 = 127.
The smallest negative 7 bit number is (11111111)b = -128
(because 11111111 is the two's complement of 10000000 = 2^7 = 128).
Unsigned chars don't have signs so they can use all the 8 bits. Going from (00000000)b = 0 to (11111111)b = 255.
Signed numbers are those that have either + or - appended with them.
E.g +2 and -6 are signed numbers.
Signed Numbers can store both positive and negative numbers thats why they have bigger range.
i.e -32768 to 32767
Unsigned numbers are simply numbers with no sign with them. they are always positive. and their range is from 0 to 65535.
Hope it helps
Signed usually means the number has a + or - symbol in front of it. This means that unsigned int, unsigned shorts, etc cannot be negative.
Nobody mentioned this, but range of int in table is wrong:
it is
-2^(31) to 2^(31)-1
i.e.,
-2,147,483,648 to 2,147,483,647
A signed integer can have both negative and positive values. While a unsigned integer can only have positive values.
For signed integers using two's complement , which is most commonly used, the range is (depending on the bit width of the integer):
char s -> range -128-127
Where a unsigned char have the range:
unsigned char s -> range 0-255
First, your table is wrong... negative numbers are missing. Refering to the type char.... you can represent at all 256 possibilities as char has one byte means 2^8. So now you have two alternatives to set ur range. either from -128 to +128 or 0 to 255. The first one is a signed char the second a unsigned char. If you using integers be aware what kind of operation system u are using. 16 bit ,32 bit or 64 bit. Int (16 bit,32 bit,64 bit). char has always just 8 bit value.
It means that there will likely be a sign ( a symbol) in front of your value (+12345 || -12345 )
Why the range of signed character is -128 to 127 but not -127 to 128 ?
That is because of the way two's complement encoding works: 0 is treated as a "positive" number (signed bit off), so, therefore, the number of available positive values is reduced by one.
In ones' complement encoding (which is not very common nowadays, but in the olden days, it was), there were separate values for +0 and -0, and so the range for an 8-bit quantity is -127 to +127.
In 8-bit 2's complement encoding numbers -128 and +128 have the same representation: 10000000. So, the designer of the hardware is presented with an obvious dilemma: how to interpret bit-pattern 10000000. Formally, it will work either way. If they decide to interpret it as +128, the resultant range will be -127..+128. If they decide to interpret it as -128, the resultant range will be -128..+127.
In actual real-life 2's complement representation the latter approach is chosen because it satisfies the following nice convention: all bit-patterns with 1 in higher-order bit represent negative numbers.
It is worth noting though, that language specification does not require 2's-complement implementations to treat the 100...0 bit pattern as a valid value in any signed integer type. E.g. implementations are allowed to restrict 8-bit signed char to -127..+127 range and regard 10000000 as an invalid bit combination (trap representation).
I think an easy way to explain this for the common soul is :
A bit is a value 0 or 1, or 2 possibilities
A 2-bit holds two combinations or 0 and 1 for four possible values : 00, 01, 10, and 11.
A 3-bit holds three combinations for a total of eight possible values : 000 to 111.
Thus n-bits holds n combinations for a total of 2^n possible values. Therefore, an 8-bit value is 2^8 = 256 possible values.
For signed numbers, the most significant bit (the first one reading the value from left to right) is the sign bit; that leaves a possibility of 2^(n-1) possible values. For an 8-bit signed number, this is 2^7 = 128 possible values for each sign. But since the positive sign includes the zero (0 to 127 = 128 different values, and 128 + 128 = 2^8 = 256), the negative sign includes -1 to... -128 for 128 different values also. Where :
10000000 = -128
...
11111111 = -1
00000000 = 0
...
01111111 = 127
#include <limits.h>
#include <stdio.h>
...
printf("range of signed character is %i ... %i", CHAR_MIN, CHAR_MAX );
If you just consider twos complement as arithmetic modulo 256, then the cutoff between positive and negative is purely arbitrary. You could just as well have put it at 63/-192, 254/-1, 130/-125, or anywhere else. However, as a standard signed integer format, twos complement came by convention put put the cutoff at 127/-128. This cutoff has one big benefit: the high bit being set corresponds directly to the number being negative.
As for the C language, it leaves the format of signed numbers up to the implementation, but only offers 3 choices of implementation, all of which use a "sign bit": sign/magnitude, ones complement, and twos complement.
If you look at ranges of chars and ints there seems to be one extra number on the negative side. This is because a negative number is always stored as 2’s compliment of its binary. For example, let us see how -128 is stored. Firstly, binary of 128 is calculated (10000000), then its 1’s compliment is obtained (01111111). A 1’s compliment is obtained by changing all 0s to 1s and 1s to 0s. Finally, 2’s compliment of this number, i.e. 10000000, gets stored. A 2’s compliment is obtained by adding 1 to the 1’s compliment. Thus, for -128, 10000000 gets stored. This is an 8-bit number and it can be easily accommodated in a char. As against this, +128 cannot be stored in a char because its binary 010000000 (left-most 0 is for positive sign) is a 9-bit number. However +127 can be stored as its binary 01111111 turns out to be a 8-bit number.
Step 1:
If you take 2's complement of any number from 0 up to 127 the bit number 8 will always be 1. So lets reserve that info.
Step2 :
if you find the -127 by applying 2's complement into +127 you will find "1 0 0 0 0 0 0 1" and finally if you substract 1 from this number then the smallest 8 bit number -128 will be achieved as "1 0 0 0 0 0 0 0"
As a result if we combine the info that we reserved at step 1 and the result from step2, we come to the conclusion that, the most significiant bit or bit number 8 in char containers must always be 1 so called signed bit.