I have a question about coverting number with IEEE 754 standart.
But I have 24 bits(1 bit is allocated to the sign / 7 bits are allocated to the exponent / remaining 16 bits are allocated to the mantissa)
Unfortunately I don't know how I can do that.
Maybe someone could help?
If you are following the ordinary IEEE-754 scheme for binary formats:
Set the sign: Starting with a number x, if x is non-negative, set the sign bit to 0. If x is negative, or you wish to encode a negative zero, set the sign bit to 1. Set t to |x| (the absolute value of x).
Determine the exponent: Set e, an exponent counter, to 0. While t is less than 1 and e is greater than −62, multiply t by 2 and add −1 to e. While t is 2 or greater and e is less than 63, multiply t by ½ and add +1 to e. (At most one of those loops will execute, depending on the starting value of t.) (The limits on the represented exponent are −62 and +63, inclusive, so the loops stop if t reaches either one.)
Round the significand: Multiply t by 224. Round it to an integer using the desired method (most often round-to-nearest, ties-to-even-low-bit). If t is 225 or greater, multiply it by ½ and add +1 to e.
If t is less than 224, x is in the subnormal (or zero) range. Set the exponent field to all zeros. Write t in binary, with leading 0 bits to make 24 bits total. Set the primary significand field to those bits. Stop.
If e is greater than 63, x does not fit in the finite range. Set the exponent field to all ones and the primary significand field to all zeros. This represents an infinity, with the sign bit indicating +∞ or −∞. Stop.
The exponent is encoded by putting e+63 in the exponent field, so write e+63 in binary, using leading 0 bits as necessary to make 7 bits, and put them into the exponent field. Write t in binary. It will be 25 bits. Put the low 24 bits in the primary significand field. Stop.
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I am told that a double in C++ has a mantissa that is capable of safely and accurately representing [-(253 − 1), 253 − 1] integers.
How is this possible when the mantissa is only 52 bits? Why is it that an int16 is only capable of having a range of [-32,768, +32,767], or [-215, 215-1], when the same thing could be used for int16 to allow twice as many representable numbers?
The format of the double (64 bits) is as follows:
1 bit: sign
11 bits: exponent
52 bits: mantissa
We only need to look at the positive integers we can represent with the mantissa, since the sign bit takes care of the negative integers for us.
Naively, with 52 bits, we can store unsigned integers from 0 to 2^52 - 1. With a sign bit, this lets us store from -2^52 - 1 to 2^52 - 1.
However, we have a little trick we can use. We say that the first digit of our integer is always a 1, which gives us an extra bit to work with.
To see why this works, let's dig a little deeper.
Every positive integer will have at least one 1 in its binary representation. So, we shift the mantissa left or right until we get a 1 at the start, using the exponent. An example might help here:
9, represented as an unsigned int: 000...0001001 (dots representing more 0s).
Written another way: 1.001 * 2^3. (1.001 being in binary, not decimal.)
And, we'll agree to never use a 0 as the first bit. So even though we could write 9 as 0.1001 * 2^4 or 0.01001 * 2^5, we won't. We'll agree that when we write the numbers out in this format, we'll always make sure we use the exponent to "shift" bits over until we start with a 1.
So, the information we need to store to get 9 is as follows:
e: 3
i: 1.001
But if i always starts with 1, why bother writing it out every time? Let's just keep the following instead:
e: 3
i: 001
Using precisely this information, we can reconstruct the number as: 1.i * 2^e == 9.
When we get up to larger numbers, our "i" will be bigger, maybe up to 52 bits used, but we're actually storing 53 bits of information because of the leading 1 we always have.
Final Note: This is not quite what is stored in the exponent and mantissa of a double, I've simplified things to help explain, but hopefully this helps people understand where the missing bit is from. Also, this does not cover 0, which gets a special representation (the trick used above will not work for 0, since the usual representation of 0 doesn't have any 1s in it).
The binary representation of d=9.2 should be something like 9.199999999999999289457264239899814128875732421875.
So 10d should be 91.999999999xxxxxxx.
Let int a = 10d, isn't it truncated as 91 instead of 92?
When the multiplication is performed, the real-number result is rounded to fit in the IEEE-754 binary64 format, and the result after rounding is exactly 92.
When 9.2 is converted to the nearest value representable in binary64, the result is indeed 9.199999999999999289457264239899814128875732421875. Using hexadecimal for the significand, this is 1.266666666666616•23.
When we multiply this by 10 using real-number arithmetic, the result is
B.7FFFFFFFFFFFC16•23. (This is easy to see. Starting from the right, 6 times 10 is 6010, which is 3C16. So we write the C digit and carry the 3. In the next column, 6 times 10 is again 3C16, and adding the carried 3 gives 3F16. We write the F digit and carry the 3. Continuing produces more F digits and propagates a carry of 3 up to the front, where 2 times 10 plus a carry of 3 is 2310 = 1716, so we write the 7 and carry the 1. Finally 1 times 10 plus 1 is 1110 = B16.)
Let’s adjust that number to normalize it to have a single bit left of the radix point. Shifting the significand right three bits and compensating by adding three to the exponent gives 1.6FFFFFFFFFFFF816•26. Now we can see this number has too many bits to fit in the binary64 format. The initial 1 has one bit, the next 13 digits have four bits, and the final 8 has one significant bit (10002, and the trailing zeros are mere placeholders, not significant bits). That is 54 bits, but the binary64 format has only 53 for the significand (52 are stored explicitly in the significand field, one is encoded via the exponent field). So we must round it to fit in the format.
The two nearest representable values are 1.6FFFFFFFFFFFF16•26 and 1.700000000000016•26. They are equally distant from 1.6FFFFFFFFFFFF816•26, and the rule for breaking ties is to choose the number with the even low bit. So the result is 1.700000000000016•26.
Thus, multiplying 9.199999999999999289457264239899814128875732421875 by 10 in binary64 arithmetic yields 1.716•26, which is 92.
I've been studying IEEE 754 for a while,and there's a thing that I do not manage to understand. According to my notes, in IEEE simple precision, you have 1 bit for the sign, 8 for exponent and 23 for mantissa, making a total of 32 bits. The exponent could be described as following: the first bit gives the sign, and then the remaining 7 bits describe some number, which means that the biggest possible value for exponent is 2^+127, and the lowest 2^-127. But according to Wikipedia (and other websites), the lowest possible value is -126 which you get if you consider the exponent as a number determined by: e-127 and e is an integer between 1 and 254. Why can't e take the value 0 which will enable the exponent -127?
Look up 'subnormal' or denormalized numbers; they have a biassed exponent value of 0.
A denormal number is represented with a biased exponent of all 0 bits, which represents an exponent of −126 in single precision (not −127).
Also, there are 24 logical bits in the mantissa, but the first is always 1 so it isn't actually stored.
Signed zeros are represented by exponent and mantissa with all bits zero, and the sign bit may be 0 (positive) or 1 (negative).
In this wiki article it shows 23 bits for precision, 8 for exponent, and 1 for sign
Where is the hidden 24th bit in float type that makes (23+1) for 7 significand digits?
Floating point numbers are usually normalized. Consider, for example, scientific notation as most of us learned it in school. You always scale the exponent so there's exactly one digit before the decimal point. For example, instead of 123.456, you write 1.23456x102.
Floating point on a computer is normally handled (almost1) the same way: numbers are normalized so there's exactly one digit before the binary point (binary point since most work in binary instead of decimal). There's one difference though: in the case of binary, that means the digit before the decimal point must be a 1. Since it's always a 1, there's no real need to store that bit. To save a bit of storage in each floating point number, that 1 bit is implicit instead of being stored.
As usual, there's just a bit more to the situation than that though. The main difference is denormalized numbers. Consider, for example, if you were doing scientific notation but you could only use exponents from -99 to +99. If you wanted to store a number like, say, 1.234*10-102, you wouldn't be able to do that directly, so it would probably just get rounded down to 0.
Denormalized numbers give you a way to deal with that. Using a denormalized number, you'd store that as 0.001234*10-99. Assuming (as is normally the case on a computer) that the number of digits for the mantissa and exponent are each limited, this loses some precision, but still avoids throwing away all the precision and just calling it 0.
1 Technically, there are differences, but they make no difference to the basic understanding involved.
http://en.wikipedia.org/wiki/Single_precision_floating-point_format#IEEE_754_single_precision_binary_floating-point_format:_binary32
The true significand includes 23
fraction bits to the right of the
binary point and an implicit leading
bit (to the left of the binary point)
with value 1 unless the exponent is
stored with all zeros
Explains it pretty well, it is by convention/design that last bit is not stored explicitly but rather stated by specification that it is there unless everything is 0'os.
As you write, the single-precision floating-point format has a sign bit, eight exponent bits, and 23 significand bits. Let s be the sign bit, e be the exponent bits, and f be the significand bits. Here is what various combinations of bits stand for:
If e and f are zero, the object is +0 or -0, according to whether s is 0 or 1.
If e is zero and f is not, the object is (-1)s * 21-127 * 0.f. "0.f" means to write 0, period, and the 23 bits of f, then interpret that as a binary numeral. E.g., 0.011000... is 3/8. These are the "subnormal" numbers.
If 0 < e < 255, the object is (-1)s * 2e-127 * 1.f. "1.f" is similar to "0.f" above, except you start with 1 instead of 0. This is the implicit bit. Most of the floating-point numbers are in this format; these are the "normal" numbers.
If e is 255 and f is zero, the object is +infinity or -infinity, according to whether s is 0 or 1.
If e is 255 and f is not zero, the object is a NaN (Not a Number). The meaning of the f field of a NaN is implementation dependent; it is not fully specified by the standard. Commonly, if the first bit is zero, it is a signaling NaN; otherwise it is a quiet NaN.
I found myself today doing some bit manipulation and I decided to refresh my floating-point knowledge a little!
Things were going great until I saw this:
... 23 fraction bits of the significand appear in the memory format but the total precision is 24 bits
I read it again and again but I still can't figure out where the 24th bit is, I noticed something about a binary point so I assumed that it's a point in the middle between the mantissa and the exponent.
I'm not really sure but I believe he author was talking about this bit:
Binary point?
|
s------e-----|-------------m----------
0 - 01111100 - 01000000000000000000000
^ this
The 24th bit is implicit due to normalization.
The significand is shifted left (and one subtracted from the exponent for each bit shift) until the leading bit of the significand is a 1.
Then, since the leading bit is a 1, only the other 23 bits are actually stored.
There is also the possibility of a denormal number. The exponent is stored as a "bias" format signed number, meaning that it's an unsigned number where the middle of the range is defined to mean 01. So, with 8 bits, it's stored as a number from 0..255, but 0 is interpreted to mean -128, 128 is interpreted to mean 0, and 255 is interpreted as 127 (I may have a fencepost error there, but you get the idea).
If, in the process of normalization, this is decremented to 0 (meaning an actual exponent value of -128), then normalization stops, and the significand is stored as-is. In this case, the implicit bit from normalization it taken to be a 0 instead of a 1.
Most floating point hardware is designed to basically assume numbers will be normalized, so they assume that implicit bit is a 1. During the computation, they check for the possibility of a denormal number, and in that case they do roughly the equivalent of throwing an exception, and re-start the calculation with that taken into account. This is why computation with denormals often gets drastically slower than otherwise.
In case you wonder why it uses this strange format: IEEE floating point (like many others) is designed to ensure that if you treat its bit pattern as an integer of the same size, you can compare them as signed, 2's complement integers and they'll still sort into the correct order as floating point numbers. Since the sign of the number is in the most significant bit (where it is for a 2's complement integer) that's treated as the sign bit. The bits of the exponent are stored as the next most significant bits -- but if we used 2's complement for them, an exponent less than 0 would set the second most significant bit of the number, which would result in what looked like a big number as an integer. By using bias format, a smaller exponent leaves that bit clear, and a larger exponent sets it, so the order as an integer reflects the order as a floating point.
Normally (pardon the pun), the leading bit of a floating point number is always 1; thus, it doesn't need to be stored anywhere. The reason is that, if it weren't 1, that would mean you had chosen the wrong exponent to represent it; you could get more precision by shifting the mantissa bits left and using a smaller exponent.
The one exception is denormal/subnormal numbers, which are represented by all zero bits in the exponent field (the lowest possible exponent). In this case, there is no implicit leading 1 in the mantissa, and you have diminishing precision as the value approaches zero.
For normal floating point numbers, the number stored in the floating point variable is (ignoring sign) 1. mantissa * 2exponent-offset. The leading 1 is not stored in the variable.