How are the values 515 and -515 (base 10) represented in 16 bit excess notation? How are they? - subscription

How are the values 515 10 and -51510 represented in 16 bit excess notation?

From the details here:
Excess-N notation is simply a way of representing number where N is the zero point. You simply subtract the zero point from the encoded numbers to get the real numbers.
For a 16-bit excess representation, the zero point is 215 or 32768.
Therefore, 515 would be represented as 32768 + 515, which is 3328310, 0x8203 or binary 1000 0010 0000 0011.
Similarly, -515 would be represented as 32768 - 515, which is 3325310, 0x7dfd or binary 0111 1101 1111 1101.

Related

Representation of negative numbers in binary

I am styding about the different data types in C. In my book it is written that the signed char data type is of
1 byte(8 bits). Now,it is written that the range of ASCII codes that can be stored in signed char data type is from -128 to 127.
It was written that a negative number is stored as a
2's complement of its binary.
I can't understand how is it possible to store the 2's complement form of -128 within 8 bits?
First of all we need 9 bits to write the signed binary representation of +128 which is 010000000.Now, if we take the 2's complement form of it, we get
110000000 which is also 9 bits long.
Then how are we able to store -128 in a 8 bit signed char?
You can store -128 as 1000 0000.
The first 1 indicates that you are using a negative number and then, how I learned it, you count the 0's as if they are 1's and vice versa and at the end you add 1. This means that for your case you would perform the following steps (when the first digit is a 1):
Take the signed char: 1000 0000.
Strip the first digit and switch the other digits: 111 1111.
Calculate, in 'simple' binary the value: in this case 127.
Add 1 and afterwards add negative sign: -128.
It is important that you take into account that all negative numbers are counted 'inversed':
-1 in decimal is written as 1111 1111
-2 in decimal is written as 1111 1110
-3 in decimal is written as 1111 1101
-4 in decimal is written as 1111 1100
etc.
This means that you have 128 possible values, but since you do not have to include the decimal 0 (because it is represented as 0000 0000), your negative range is [-1, -128], while on the other hand the range is [0, 127].

-2^31 as smallest integer, why? [duplicate]

I'm in a computer systems course and have been struggling, in part, with two's complement. I want to understand it, but everything I've read hasn't brought the picture together for me. I've read the Wikipedia article and various other articles, including my text book.
What is two's complement, how can we use it and how can it affect numbers during operations like casts (from signed to unsigned and vice versa), bit-wise operations and bit-shift operations?
Two's complement is a clever way of storing integers so that common math problems are very simple to implement.
To understand, you have to think of the numbers in binary.
It basically says,
for zero, use all 0's.
for positive integers, start counting up, with a maximum of 2(number of bits - 1)-1.
for negative integers, do exactly the same thing, but switch the role of 0's and 1's and count down (so instead of starting with 0000, start with 1111 - that's the "complement" part).
Let's try it with a mini-byte of 4 bits (we'll call it a nibble - 1/2 a byte).
0000 - zero
0001 - one
0010 - two
0011 - three
0100 to 0111 - four to seven
That's as far as we can go in positives. 23-1 = 7.
For negatives:
1111 - negative one
1110 - negative two
1101 - negative three
1100 to 1000 - negative four to negative eight
Note that you get one extra value for negatives (1000 = -8) that you don't for positives. This is because 0000 is used for zero. This can be considered as Number Line of computers.
Distinguishing between positive and negative numbers
Doing this, the first bit gets the role of the "sign" bit, as it can be used to distinguish between nonnegative and negative decimal values. If the most significant bit is 1, then the binary can be said to be negative, where as if the most significant bit (the leftmost) is 0, you can say the decimal value is nonnegative.
"Sign-magnitude" negative numbers just have the sign bit flipped of their positive counterparts, but this approach has to deal with interpreting 1000 (one 1 followed by all 0s) as "negative zero" which is confusing.
"Ones' complement" negative numbers are just the bit-complement of their positive counterparts, which also leads to a confusing "negative zero" with 1111 (all ones).
You will likely not have to deal with Ones' Complement or Sign-Magnitude integer representations unless you are working very close to the hardware.
I wonder if it could be explained any better than the Wikipedia article.
The basic problem that you are trying to solve with two's complement representation is the problem of storing negative integers.
First, consider an unsigned integer stored in 4 bits. You can have the following
0000 = 0
0001 = 1
0010 = 2
...
1111 = 15
These are unsigned because there is no indication of whether they are negative or positive.
Sign Magnitude and Excess Notation
To store negative numbers you can try a number of things. First, you can use sign magnitude notation which assigns the first bit as a sign bit to represent +/- and the remaining bits to represent the magnitude. So using 4 bits again and assuming that 1 means - and 0 means + then you have
0000 = +0
0001 = +1
0010 = +2
...
1000 = -0
1001 = -1
1111 = -7
So, you see the problem there? We have positive and negative 0. The bigger problem is adding and subtracting binary numbers. The circuits to add and subtract using sign magnitude will be very complex.
What is
0010
1001 +
----
?
Another system is excess notation. You can store negative numbers, you get rid of the two zeros problem but addition and subtraction remains difficult.
So along comes two's complement. Now you can store positive and negative integers and perform arithmetic with relative ease. There are a number of methods to convert a number into two's complement. Here's one.
Convert Decimal to Two's Complement
Convert the number to binary (ignore the sign for now)
e.g. 5 is 0101 and -5 is 0101
If the number is a positive number then you are done.
e.g. 5 is 0101 in binary using two's complement notation.
If the number is negative then
3.1 find the complement (invert 0's and 1's)
e.g. -5 is 0101 so finding the complement is 1010
3.2 Add 1 to the complement 1010 + 1 = 1011.
Therefore, -5 in two's complement is 1011.
So, what if you wanted to do 2 + (-3) in binary? 2 + (-3) is -1.
What would you have to do if you were using sign magnitude to add these numbers? 0010 + 1101 = ?
Using two's complement consider how easy it would be.
2 = 0010
-3 = 1101 +
-------------
-1 = 1111
Converting Two's Complement to Decimal
Converting 1111 to decimal:
The number starts with 1, so it's negative, so we find the complement of 1111, which is 0000.
Add 1 to 0000, and we obtain 0001.
Convert 0001 to decimal, which is 1.
Apply the sign = -1.
Tada!
Like most explanations I've seen, the ones above are clear about how to work with 2's complement, but don't really explain what they are mathematically. I'll try to do that, for integers at least, and I'll cover some background that's probably familiar first.
Recall how it works for decimal: 2345 is a way of writing 2 × 103 + 3 × 102 + 4 × 101 + 5 × 100.
In the same way, binary is a way of writing numbers using just 0 and 1 following the same general idea, but replacing those 10s above with 2s. Then in binary, 1111is a way of writing 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20and if you work it out, that turns out to equal 15 (base 10). That's because it is 8+4+2+1 = 15.
This is all well and good for positive numbers. It even works for negative numbers if you're willing to just stick a minus sign in front of them, as humans do with decimal numbers. That can even be done in computers, sort of, but I haven't seen such a computer since the early 1970's. I'll leave the reasons for a different discussion.
For computers it turns out to be more efficient to use a complement representation for negative numbers. And here's something that is often overlooked. Complement notations involve some kind of reversal of the digits of the number, even the implied zeroes that come before a normal positive number. That's awkward, because the question arises: all of them? That could be an infinite number of digits to be considered.
Fortunately, computers don't represent infinities. Numbers are constrained to a particular length (or width, if you prefer). So let's return to positive binary numbers, but with a particular size. I'll use 8 digits ("bits") for these examples. So our binary number would really be 00001111or 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20
To form the 2's complement negative, we first complement all the (binary) digits to form 11110000and add 1 to form 11110001but how are we to understand that to mean -15?
The answer is that we change the meaning of the high-order bit (the leftmost one). This bit will be a 1 for all negative numbers. The change will be to change the sign of its contribution to the value of the number it appears in. So now our 11110001 is understood to represent -1 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20Notice that "-" in front of that expression? It means that the sign bit carries the weight -27, that is -128 (base 10). All the other positions retain the same weight they had in unsigned binary numbers.
Working out our -15, it is -128 + 64 + 32 + 16 + 1 Try it on your calculator. it's -15.
Of the three main ways that I've seen negative numbers represented in computers, 2's complement wins hands down for convenience in general use. It has an oddity, though. Since it's binary, there have to be an even number of possible bit combinations. Each positive number can be paired with its negative, but there's only one zero. Negating a zero gets you zero. So there's one more combination, the number with 1 in the sign bit and 0 everywhere else. The corresponding positive number would not fit in the number of bits being used.
What's even more odd about this number is that if you try to form its positive by complementing and adding one, you get the same negative number back. It seems natural that zero would do this, but this is unexpected and not at all the behavior we're used to because computers aside, we generally think of an unlimited supply of digits, not this fixed-length arithmetic.
This is like the tip of an iceberg of oddities. There's more lying in wait below the surface, but that's enough for this discussion. You could probably find more if you research "overflow" for fixed-point arithmetic. If you really want to get into it, you might also research "modular arithmetic".
2's complement is very useful for finding the value of a binary, however I thought of a much more concise way of solving such a problem(never seen anyone else publish it):
take a binary, for example: 1101 which is [assuming that space "1" is the sign] equal to -3.
using 2's complement we would do this...flip 1101 to 0010...add 0001 + 0010 ===> gives us 0011. 0011 in positive binary = 3. therefore 1101 = -3!
What I realized:
instead of all the flipping and adding, you can just do the basic method for solving for a positive binary(lets say 0101) is (23 * 0) + (22 * 1) + (21 * 0) + (20 * 1) = 5.
Do exactly the same concept with a negative!(with a small twist)
take 1101, for example:
for the first number instead of 23 * 1 = 8 , do -(23 * 1) = -8.
then continue as usual, doing -8 + (22 * 1) + (21 * 0) + (20 * 1) = -3
Imagine that you have a finite number of bits/trits/digits/whatever. You define 0 as all digits being 0, and count upwards naturally:
00
01
02
..
Eventually you will overflow.
98
99
00
We have two digits and can represent all numbers from 0 to 100. All those numbers are positive! Suppose we want to represent negative numbers too?
What we really have is a cycle. The number before 2 is 1. The number before 1 is 0. The number before 0 is... 99.
So, for simplicity, let's say that any number over 50 is negative. "0" through "49" represent 0 through 49. "99" is -1, "98" is -2, ... "50" is -50.
This representation is ten's complement. Computers typically use two's complement, which is the same except using bits instead of digits.
The nice thing about ten's complement is that addition just works. You do not need to do anything special to add positive and negative numbers!
I read a fantastic explanation on Reddit by jng, using the odometer as an analogy.
It is a useful convention. The same circuits and logic operations that
add / subtract positive numbers in binary still work on both positive
and negative numbers if using the convention, that's why it's so
useful and omnipresent.
Imagine the odometer of a car, it rolls around at (say) 99999. If you
increment 00000 you get 00001. If you decrement 00000, you get 99999
(due to the roll-around). If you add one back to 99999 it goes back to
00000. So it's useful to decide that 99999 represents -1. Likewise, it is very useful to decide that 99998 represents -2, and so on. You have
to stop somewhere, and also by convention, the top half of the numbers
are deemed to be negative (50000-99999), and the bottom half positive
just stand for themselves (00000-49999). As a result, the top digit
being 5-9 means the represented number is negative, and it being 0-4
means the represented is positive - exactly the same as the top bit
representing sign in a two's complement binary number.
Understanding this was hard for me too. Once I got it and went back to
re-read the books articles and explanations (there was no internet
back then), it turned out a lot of those describing it didn't really
understand it. I did write a book teaching assembly language after
that (which did sell quite well for 10 years).
Two complement is found out by adding one to 1'st complement of the given number.
Lets say we have to find out twos complement of 10101 then find its ones complement, that is, 01010 add 1 to this result, that is, 01010+1=01011, which is the final answer.
Lets get the answer 10 – 12 in binary form using 8 bits:
What we will really do is 10 + (-12)
We need to get the compliment part of 12 to subtract it from 10.
12 in binary is 00001100.
10 in binary is 00001010.
To get the compliment part of 12 we just reverse all the bits then add 1.
12 in binary reversed is 11110011. This is also the Inverse code (one's complement).
Now we need to add one, which is now 11110100.
So 11110100 is the compliment of 12! Easy when you think of it this way.
Now you can solve the above question of 10 - 12 in binary form.
00001010
11110100
-----------------
11111110
Looking at the two's complement system from a math point of view it really makes sense. In ten's complement, the idea is to essentially 'isolate' the difference.
Example: 63 - 24 = x
We add the complement of 24 which is really just (100 - 24). So really, all we are doing is adding 100 on both sides of the equation.
Now the equation is: 100 + 63 - 24 = x + 100, that is why we remove the 100 (or 10 or 1000 or whatever).
Due to the inconvenient situation of having to subtract one number from a long chain of zeroes, we use a 'diminished radix complement' system, in the decimal system, nine's complement.
When we are presented with a number subtracted from a big chain of nines, we just need to reverse the numbers.
Example: 99999 - 03275 = 96724
That is the reason, after nine's complement, we add 1. As you probably know from childhood math, 9 becomes 10 by 'stealing' 1. So basically it's just ten's complement that takes 1 from the difference.
In Binary, two's complement is equatable to ten's complement, while one's complement to nine's complement. The primary difference is that instead of trying to isolate the difference with powers of ten (adding 10, 100, etc. into the equation) we are trying to isolate the difference with powers of two.
It is for this reason that we invert the bits. Just like how our minuend is a chain of nines in decimal, our minuend is a chain of ones in binary.
Example: 111111 - 101001 = 010110
Because chains of ones are 1 below a nice power of two, they 'steal' 1 from the difference like nine's do in decimal.
When we are using negative binary number's, we are really just saying:
0000 - 0101 = x
1111 - 0101 = 1010
1111 + 0000 - 0101 = x + 1111
In order to 'isolate' x, we need to add 1 because 1111 is one away from 10000 and we remove the leading 1 because we just added it to the original difference.
1111 + 1 + 0000 - 0101 = x + 1111 + 1
10000 + 0000 - 0101 = x + 10000
Just remove 10000 from both sides to get x, it's basic algebra.
The word complement derives from completeness. In the decimal world the numerals 0 through 9 provide a complement (complete set) of numerals or numeric symbols to express all decimal numbers. In the binary world the numerals 0 and 1 provide a complement of numerals to express all binary numbers. In fact The symbols 0 and 1 must be used to represent everything (text, images, etc) as well as positive (0) and negative (1).
In our world the blank space to the left of number is considered as zero:
35=035=000000035.
In a computer storage location there is no blank space. All bits (binary digits) must be either 0 or 1. To efficiently use memory numbers may be stored as 8 bit, 16 bit, 32 bit, 64 bit, 128 bit representations. When a number that is stored as an 8 bit number is transferred to a 16 bit location the sign and magnitude (absolute value) must remain the same. Both 1's complement and 2's complement representations facilitate this.
As a noun:
Both 1's complement and 2's complement are binary representations of signed quantities where the most significant bit (the one on the left) is the sign bit. 0 is for positive and 1 is for negative.
2s complement does not mean negative. It means a signed quantity. As in decimal the magnitude is represented as the positive quantity. The structure uses sign extension to preserve the quantity when promoting to a register [] with more bits:
[0101]=[00101]=[00000000000101]=5 (base 10)
[1011]=[11011]=[11111111111011]=-5(base 10)
As a verb:
2's complement means to negate. It does not mean make negative. It means if negative make positive; if positive make negative. The magnitude is the absolute value:
if a >= 0 then |a| = a
if a < 0 then |a| = -a = 2scomplement of a
This ability allows efficient binary subtraction using negate then add.
a - b = a + (-b)
The official way to take the 1's complement is for each digit subtract its value from 1.
1'scomp(0101) = 1010.
This is the same as flipping or inverting each bit individually. This results in a negative zero which is not well loved so adding one to te 1's complement gets rid of the problem.
To negate or take the 2s complement first take the 1s complement then add 1.
Example 1 Example 2
0101 --original number 1101
1's comp 1010 0010
add 1 0001 0001
2's comp 1011 --negated number 0011
In the examples the negation works as well with sign extended numbers.
Adding:
1110 Carry 111110 Carry
0110 is the same as 000110
1111 111111
sum 0101 sum 000101
SUbtracting:
1110 Carry 00000 Carry
0110 is the same as 00110
-0111 +11001
---------- ----------
sum 0101 sum 11111
Notice that when working with 2's complement, blank space to the left of the number is filled with zeros for positive numbers butis filled with ones for negative numbers. The carry is always added and must be either a 1 or 0.
Cheers
2's complement is essentially a way of coming up with the additive inverse of a binary number. Ask yourself this: Given a number in binary form (present at a fixed length memory location), what bit pattern, when added to the original number (at the fixed length memory location), would make the result all zeros ? (at the same fixed length memory location). If we could come up with this bit pattern then that bit pattern would be the -ve representation (additive inverse) of the original number; as by definition adding a number to its additive inverse always results in zero. Example: take 5 which is 101 present inside a single 8 bit byte. Now the task is to come up with a bit pattern which when added to the given bit pattern (00000101) would result in all zeros at the memory location which is used to hold this 5 i.e. all 8 bits of the byte should be zero. To do that, start from the right most bit of 101 and for each individual bit, again ask the same question: What bit should I add to the current bit to make the result zero ? continue doing that taking in account the usual carry over. After we are done with the 3 right most places (the digits that define the original number without regard to the leading zeros) the last carry goes in the bit pattern of the additive inverse. Furthermore, since we are holding in the original number in a single 8 bit byte, all other leading bits in the additive inverse should also be 1's so that (and this is important) when the computer adds "the number" (represented using the 8 bit pattern) and its additive inverse using "that" storage type (a byte) the result in that byte would be all zeros.
1 1 1
----------
1 0 1
1 0 1 1 ---> additive inverse
---------
0 0 0
Many of the answers so far nicely explain why two's complement is used to represent negative numbers, but do not tell us what two's complement number is, particularly not why a '1' is added, and in fact often added in a wrong way.
The confusion comes from a poor understanding of the definition of a complement number. A complement is the missing part that would make something complete.
The radix complement of an n digit number x in radix b is, by definition, b^n-x.
In binary 4 is represented by 100, which has 3 digits (n=3) and a radix of 2 (b=2). So its radix complement is b^n-x = 2^3-4=8-4=4 (or 100 in binary).
However, in binary obtaining a radix's complement is not as easy as getting its diminished radix complement, which is defined as (b^n-1)-y, just 1 less than that of radix complement. To get a diminished radix complement, you simply flip all the digits.
100 -> 011 (diminished (one's) radix complement)
to obtain the radix (two's) complement, we simply add 1, as the definition defined.
011 +1 ->100 (two's complement).
Now with this new understanding, let's take a look of the example given by Vincent Ramdhanie (see above second response):
Converting 1111 to decimal:
The number starts with 1, so it's negative, so we find the complement of 1111, which is 0000.
Add 1 to 0000, and we obtain 0001.
Convert 0001 to decimal, which is 1.
Apply the sign = -1.
Tada!
Should be understood as:
The number starts with 1, so it's negative. So we know it is a two's complement of some value x. To find the x represented by its two's complement, we first need find its 1's complement.
two's complement of x: 1111
one's complement of x: 1111-1 ->1110;
x = 0001, (flip all digits)
Apply the sign -, and the answer =-x =-1.
I liked lavinio's answer, but shifting bits adds some complexity. Often there's a choice of moving bits while respecting the sign bit or while not respecting the sign bit. This is the choice between treating the numbers as signed (-8 to 7 for a nibble, -128 to 127 for bytes) or full-range unsigned numbers (0 to 15 for nibbles, 0 to 255 for bytes).
It is a clever means of encoding negative integers in such a way that approximately half of the combination of bits of a data type are reserved for negative integers, and the addition of most of the negative integers with their corresponding positive integers results in a carry overflow that leaves the result to be binary zero.
So, in 2's complement if one is 0x0001 then -1 is 0x1111, because that will result in a combined sum of 0x0000 (with an overflow of 1).
2’s Complements: When we add an extra one with the 1’s complements of a number we will get the 2’s complements. For example: 100101 it’s 1’s complement is 011010 and 2’s complement is 011010+1 = 011011 (By adding one with 1's complement) For more information
this article explain it graphically.
Two's complement is mainly used for the following reasons:
To avoid multiple representations of 0
To avoid keeping track of carry bit (as in one's complement) in case of an overflow.
Carrying out simple operations like addition and subtraction becomes easy.
Two's complement is one of the ways of expressing a negative number and most of the controllers and processors store a negative number in two's complement form.
In simple terms, two's complement is a way to store negative numbers in computer memory. Whereas positive numbers are stored as a normal binary number.
Let's consider this example,
The computer uses the binary number system to represent any number.
x = 5;
This is represented as 0101.
x = -5;
When the computer encounters the - sign, it computes its two's complement and stores it.
That is, 5 = 0101 and its two's complement is 1011.
The important rules the computer uses to process numbers are,
If the first bit is 1 then it must be a negative number.
If all the bits except first bit are 0 then it is a positive number, because there is no -0 in number system (1000 is not -0 instead it is positive 8).
If all the bits are 0 then it is 0.
Else it is a positive number.
To bitwise complement a number is to flip all the bits in it. To two’s complement it, we flip all the bits and add one.
Using 2’s complement representation for signed integers, we apply the 2’s complement operation to convert a positive number to its negative equivalent and vice versa. So using nibbles for an example, 0001 (1) becomes 1111 (-1) and applying the op again, returns to 0001.
The behaviour of the operation at zero is advantageous in giving a single representation for zero without special handling of positive and negative zeroes. 0000 complements to 1111, which when 1 is added. overflows to 0000, giving us one zero, rather than a positive and a negative one.
A key advantage of this representation is that the standard addition circuits for unsigned integers produce correct results when applied to them. For example adding 1 and -1 in nibbles: 0001 + 1111, the bits overflow out of the register, leaving behind 0000.
For a gentle introduction, the wonderful Computerphile have produced a video on the subject.
The question is 'What is “two's complement”?'
The simple answer for those wanting to understand it theoretically (and me seeking to complement the other more practical answers): 2's complement is the representation for negative integers in the dual system that does not require additional characters, such as + and -.
Two's complement of a given number is the number got by adding 1 with the ones' complement of the number.
Suppose, we have a binary number: 10111001101
Its 1's complement is: 01000110010
And its two's complement will be: 01000110011
Reference: Two's Complement (Thomas Finley)
I invert all the bits and add 1. Programmatically:
// In C++11
int _powers[] = {
1,
2,
4,
8,
16,
32,
64,
128
};
int value = 3;
int n_bits = 4;
int twos_complement = (value ^ ( _powers[n_bits]-1)) + 1;
You can also use an online calculator to calculate the two's complement binary representation of a decimal number: http://www.convertforfree.com/twos-complement-calculator/
The simplest answer:
1111 + 1 = (1)0000. So 1111 must be -1. Then -1 + 1 = 0.
It's perfect to understand these all for me.

Fraction to right of radix - Floating point conversion

When converting a number from base 10 to binary using the floating point bit model, what determines how many zeros you "zero pad" the fraction to the right of the radix?
Take for example -44.375
It was a question on a test in my systems programming course, and below is the answer the prof provided the class with... I posted this because most comments below seem to argue what my professor states in the answer and causing some confusion.
Answer: 1 1000 0100 0110 0011 0000 0000 0000 000
-- sign bit: 1
-- fixed point: -44.375 = 25 + 23 + 22 + 2-2 + 2-3
= 101100.011
= 1.01100011 * 2<sup>5</sup>
-- exponent: 5 + 127 = 132 = 1000 0100
-- fraction: 0110 0011 0000 0000 0000 000
Marking:
-- 1 mark for correct sign bit
-- 2 marks for correct fixed point representation
-- 2 marks for correct exponent (in binary)
-- 2 marks for correct fraction (padded with zeros)
Unless the float is very small, there is no left "zero pad" of the fraction.
The sample here is -1.63 (in hexadecimal) * power(2,5 (decimal)).
The exponent is adjusted until the leading digit is 1.
printf("%a\n", -44.375);
// -0x1.63p+5
[Edit]
Your prof wants to see "2 marks for correct fraction (padded with zeros)" as the number of bits in a float, so the significand in your example is
1.0110 0011 0000 0000 0000 000
The leading 1 is not stored explicitly in a typical float.
OP "what determines how many zeros you "zero pad" the fraction to the right of the radix?
A: IEEE 754 binary32 (a popular float implementation) has a 24 bit significand. A lead bit (usually 1) and a 23-bit fraction. Thus your "right" zero padding goes out to fill 23 places.
To determine the significand of an IEEE-754 32-bit binary floating-point value:
Figure out where the leading (most significant) 1 bit is. That is the starting point. Calculate 23 more bits. If there is anything left over, round it into last of the 24 bits (carrying as necessary).
Exception: If the leading bit is less than 2-126, use the 2-126 bit as the starting point, even though it is zero.
That gives the mathematical significand. To get the bits for the significand field, remove the first bit. (And, if the exception was used, set the encoded exponent to zero instead of the normal value.)
Another exception: If the leading bit, after rounding, is 2128 or greater, the conversion overflows. Set the result to infinity.

Why does the range of int has a minus 1?

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.

Converting IEEE 754 Float to MIL-STD-1750A Float

I am trying to convert a IEEE 754 32 bit single precision floating point value (standard c float variable) to an unsigned long variable in the format of MIL-STD-1750A. I have included the specification for both IEEE 754 and MIL-STD-1750A at the bottom of the post. Right now, I am having issues in my code with converting the exponent. I also see issues with converting the mantissa, but I haven't gotten to fixing those yet. I am using the examples listed in Table 3 in the link above to confirm if my program is converting properly. Some of those examples do not make sense to me.
How can these two examples have the same exponent?
.5 x 2^0 (0100 0000 0000 0000 0000 0000 0000 0000)
-1 x 2^0 (1000 0000 0000 0000 0000 0000 0000 0000)
.5 x 2^0 has one decimal place, and -1 has no decimal places, so the value for .5 x 2^0 should be
.5 x 2^0 (0100 0000 0000 0000 0000 0000 0000 0010)
right? (0010 instead of 0001, because 1750A uses plus 1 bias)
How can the last example use all 32 bits and the first bit be 1, indicating a negative value?
0.7500001x2^4 (1001 1111 1111 1111 1111 1111 0000 0100)
I can see that a value with a 127 exponent should be 7F (0111 1111) but what about a value with a negative 127 exponent? Would it be 81 (1000 0001)? If so, is it because that is the two's complement +1 of 127?
Thank you
1) How can these two examples have the same exponent?
As I understand it, the sign and mantissa effectively define a 2's-complement value in the range [-1.0,1.0).
Of course, this leads to redundant representations (0.125*21 = 0.25*20, etc.) So a canonical normalized representation is chosen, by disallowing mantissa values in the range [-0.5,0.5).
So in your two examples, both -1.0 and 0.5 fall into the "allowed" mantissa range, so they both share the same exponent value.
2) How can the last example use all 32 bits and the first bit be 1, indicating a negative value?
That doesn't look right to me; how did you obtain that representation?
3) What about a value with a negative 127 exponent? Would it be 81 (1000 0001)?
I believe so.
Remember the fraction is a "signed fraction". The signed values are stored in 2's complement format. So think of the zeros as ones.
Thus the number can be written as -0.111111111111111111111 (base 2) x 2^0
, which is close to one (converges to 1.0 if my math is correct)
On the last example, there is a negative sign in the original document (-0.7500001x2^4)

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