Computers represent real values in a form similar to that of scientific notation. Consider the value
1.23 x 10^4
The number has a sign (+ in this case)
The significand (1.23) is written with one non-zero digit
to the left of the decimal point.
The base (radix) is 10.
The exponent (an integer value) is 4. It too must have a sign.
There are standards which define what the representation means, so that across computers there will be consistancy.
Note that this is not the only way to represent floating point numbers, it is just the IEEE standard way of doing it.
Here is what we do:
the representation has three fields:
---------------------------- | S | E | F | ----------------------------
the decimal value represented is:
S e (-1) x f x 2where
e = E - bias f = ( F/(2^n) ) + 1
for single precision representation (the emphasis in this class)
n = 23
bias = 127
for double precision representation (a 64-bit representation)
n = 52 (there are 52 bits for the mantissa field)
bias = 1023 (there are 11 bits for the exponent field)
Now, what does all this mean?
An example: Put the decimal number 64.2 into the IEEE standard single precision floating point representation.
first step: get a binary representation for 64.2 to do this, get unsigned binary representations for the stuff to the left and right of the decimal point separately. 64 is 1000000 .2 can be gotten using the algorithm: .2 x 2 = 0.4 0 .4 x 2 = 0.8 0 .8 x 2 = 1.6 1 .6 x 2 = 1.2 1 .2 x 2 = 0.4 0 now this whole pattern (0011) repeats. .4 x 2 = 0.8 0 .8 x 2 = 1.6 1 .6 x 2 = 1.2 1 so a binary representation for .2 is .001100110011. . . ---- or .0011 (The bar over the top shows which bits repeat.) Putting the halves back together again: 64.2 is 1000000.0011001100110011. . . second step: Normalize the binary representation. (make it look like scientific notation) 6 1.000000 00110011. . . x 2 third step: 6 is the true exponent. For the standard form, it needs to be in 8-bit, biased-127 representation. 6 + 127 ----- 133 133 in 8-bit, unsigned representation is 1000 0101 This is the bit pattern used for E in the standard form. fourth step: the mantissa stored (F) is the stuff to the right of the radix point in the normalized form. We need 23 bits of it. 000000 00110011001100110 put it all together (and include the correct sign bit): S E F 0 10000101 00000000110011001100110 the values are often given in hex, so here it is 0100 0010 1000 0000 0110 0110 0110 0110 0x 4 2 8 0 6 6 6 6
Some extra details:
We take the bit patterns 0x0000 0000 and 0x8000 0000 to represent the value 0.
(What floating point numbers cannot be represented because of this?)
Note that the hardware that does arithmetic on floating point numbers must be constantly checking to see if it needs to use a hidden bit of a 1 or a hidden bit of 0 (for 0.0).
Values that are very close to 0.0, and would require the hidden bit to be a zero are called denormalized or subnormal numbers.
S E F 0.0 0 or 1 00000000 00000000000000000000000 (hidden bit is a 0) subnormal 0 or 1 00000000 not all zeros (hidden bit is a 0) normalized 0 or 1 > 0 any bit pattern (hidden bit is a 1)
S E F +infinity 0 11111111 00000... (0x7f80 0000) -infinity 1 11111111 00000... (0xff80 0000) NaN (Not a Number) ? 11111111 ?????... (S is either 0 or 1, E=0xff, and F is anything but all zeros)
For double precision:
arithmetic operations on floating point numbers consist of addition, subtraction, multiplication and division the operations are done with algorithms similar to those used on sign magnitude integers (because of the similarity of representation) -- example, only add numbers of the same sign. If the numbers are of opposite sign, must do subtraction.
example on decimal value given in scientific notation: 3.25 x 10 ** 3 + 2.63 x 10 ** -1 ----------------- first step: align decimal points second step: add 3.25 x 10 ** 3 + 0.000263 x 10 ** 3 -------------------- 3.250263 x 10 ** 3 (presumes use of infinite precision, without regard for accuracy) third step: normalize the result (already normalized!) example on fl pt. value given in binary: S E F .25 = 0 01111101 00000000000000000000000 100 = 0 10000101 10010000000000000000000 to add these fl. pt. representations, step 1: align radix points shifting the mantissa LEFT by 1 bit DECREASES THE EXPONENT by 1 shifting the mantissa RIGHT by 1 bit INCREASES THE EXPONENT by 1 we want to shift the mantissa right, because the bits that fall off the end should come from the least significant end of the mantissa -> choose to shift the .25, since we want to increase it's exponent. -> shift by 10000101 -01111101 --------- 00001000 (8) places. with hidden bit and radix point shown, for clarity: 0 01111101 1.00000000000000000000000 (original value) 0 01111110 0.10000000000000000000000 (shifted 1 place) (note that hidden bit is shifted into msb of mantissa) 0 01111111 0.01000000000000000000000 (shifted 2 places) 0 10000000 0.00100000000000000000000 (shifted 3 places) 0 10000001 0.00010000000000000000000 (shifted 4 places) 0 10000010 0.00001000000000000000000 (shifted 5 places) 0 10000011 0.00000100000000000000000 (shifted 6 places) 0 10000100 0.00000010000000000000000 (shifted 7 places) 0 10000101 0.00000001000000000000000 (shifted 8 places) step 2: add (don't forget the hidden bit for the 100) 0 10000101 1.10010000000000000000000 (100) + 0 10000101 0.00000001000000000000000 (.25) --------------------------------------- 0 10000101 1.10010001000000000000000 step 3: normalize the result (get the "hidden bit" to be a 1) it already is for this example. result is 0 10000101 10010001000000000000000 suppose that the result of an addition of aligned mantissas gives 10.11110000000000000000000 and the exponent to go with this is 10000000. We must put the mantissa back in the normalized form. Shift the mantissa to the right by one place, and increase the exponent by 1. The exponent and mantissa become 10000001 1.01111000000000000000000 0 (1 bit is lost off the least significant end)
like addition as far as alignment of radix points then the algorithm for subtraction of sign mag. numbers takes over. before subtracting, compare magnitudes (don't forget the hidden bit!) change sign bit if order of operands is changed. don't forget to normalize number afterward. EXAMPLE: 0 10000001 10010001000000000000000 (the representations) - 0 10000000 11100000000000000000000 --------------------------------------- step 1: align radix points 0 10000000 11100000000000000000000 becomes 0 10000001 11110000000000000000000 (notice hidden bit shifted in) 0 10000001 1.10010001000000000000000 - 0 10000001 0.11110000000000000000000 --------------------------------------- step 2: subtract mantissa 1.10010001000000000000000 - 0.11110000000000000000000 ------------------------------- 0.10100001000000000000000 step 3: put result in normalized form Shift mantissa left by 1 place, implying a subtraction of 1 from the exponent. 0 10000000 01000010000000000000000
example on decimal values given in scientific notation: 3.0 x 10 ** 1 + 0.5 x 10 ** 2 ----------------- algorithm: multiply mantissas (use unsigned multiplication) add exponents 3.0 x 10 ** 1 + 0.5 x 10 ** 2 ----------------- 1.50 x 10 ** 3 example in binary: use a mantissa that is only 4 bits as an example 0 10000100 0100 x 1 00111100 1100 ----------------- The multiplication is unsigned multiplication (NOT two's complement), so, make sure that all bits are included in the answer. Do NOT do any sign extension! mantissa multiplication: 1.0100 (don't forget hidden bit) x 1.1100 ------ 00000 00000 10100 10100 10100 --------- 1000110000 becomes 10.00110000 add exponents: always add true exponents (otherwise the bias gets added in twice) biased: 10000100 + 00111100 ---------- 10000100 01111111 (switch the order of the subtraction, - 01111111 - 00111100 so that we can get a negative value) ---------- ---------- 00000101 01000011 true exp true exp is 5. is -67 add true exponents 5 + (-67) is -62. re-bias exponent: -62 + 127 is 65. unsigned representation for 65 is 01000001. put the result back together (and add sign bit). 1 01000001 10.00110000 normalize the result: (moving the radix point one place to the left increases the exponent by 1. Can also think of this as a logical right shift.) 1 01000001 10.00110000 becomes 1 01000010 1.000110000 this is the value stored (not the hidden bit!): 1 01000010 000110000
similar to multiplication. true division: do unsigned division on the mantissas (don't forget the hidden bit) subtract TRUE exponents The IEEE standard is very specific about how all this is done. Unfortunately, the hardware to do all this is pretty slow. Some comparisons of approximate times: 2's complement integer add 1 time unit fl. pt add 4 time units fl. pt multiply 6 time units fl. pt. divide 13 time units There is a faster way to do division. Its called division by reciprocal approximation. It takes about the same time as a fl. pt. multiply. Unfortunately, the results are not always correct. Division by reciprocal approximation: instead of doing a / b they do a x 1/b. figure out a reciprocal for b, and then use the fl. pt. multiplication hardware. example of a result that isn't the same as with true division. true division: 3/3 = 1 (exactly) reciprocal approx: 1/3 = .33333333 3 x .33333333 = .99999999, not 1 It is not always possible to get a perfectly accurate reciprocal. Current fastest (and correct) division algorithm is called SRT Sweeney, Robertson, and Tocher Uses redundant quotient representation E.g., base 4 usually has digits {0,1,2,3} SRT's redundant base 4 has digits {-2,-1,0,+1,+2} Allows division algorithm to guess digits approximately with a table lookup. Approximations are fixed up when less-sigificant digits are calculated Final result in completely-accurate binary. In 1994, Intel got a few corner cases of the table wrong Maximum error less than one part in 10,000 Nevertheless, Intel took a $300M write-off to replace chip Compare with software bugs that give the wrong answer and the customer pays for the upgrade
note: this discussion only touches the surface of some issues that people deal with. Entire courses are taught on floating point arithmetic. use of standards ---------------- --> allows all machines following the standard to exchange data and to calculate the exact same results. --> IEEE fl. pt. standard sets parameters of data representation (# bits for mantissa vs. exponent) --> MIPS architecture follows the standard (All architectures follow the standard now.) overflow and underflow ---------------------- Just as with integer arithmetic, floating point arithmetic operations can cause overflow. Detection of overflow in fl. pt. comes by checking exponents before/during normalization. Once overflow has occurred, an infinity value can be represented and propagated through a calculation. Underflow occurs in fl. pt. representations when a number is to small (close to 0) to be represented. (show number line!) if a fl. pt. value cannot be normalized (getting a 1 just to the left of the radix point would cause the exponent field to be all 0's) then underflow occurs. Underflow may result in the representation of denormalized values. These are ones in which the hidden bit is a 0. The exponent field will also be all 0s in this case. Note that there would be a reduced precision for representing the mantissa (less than 23 bits to the right of the radix point). Defining the results of certain operations ------------------------------------------ Many operations result in values that cannot be represented. Other operations have undefined results. Here is a list of some operations and their results as defined in the standard. 1/0 = infinity any positive value / 0 = positive infinity any negative value / 0 = negative infinity infinity * x = infinity 1/infinity = 0 0/0 = NaN 0 * infinity = NaN infinity * infinity = NaN infinity - infinity = NaN infinity/infinity = NaN x + NaN = NaN NaN + x = NaN x - NaN = NaN NaN - x = NaN sqrt(negative value) = NaN NaN * x = NaN NaN * 0 = NaN 1/NaN = NaN (Any operation on a NaN produces a NaN.) HW vs. SW computing ------------------- floating point operations can be done by hardware (circuitry) or by software (program code). -> a programmer won't know which is occuring, without prior knowledge of the HW. -> SW is much slower than HW. by approx. 1000 times. A difficult (but good) exercize for students would be to design a SW algorithm for doing fl. pt. addition using only integer operations. SW to do fl. pt. operations is tedious. It takes lots of shifting and masking to get the data in the right form to use integer arithmetic operations to get a result -- and then more shifting and masking to put the number back into fl. pt. format. A common thing for manufacturers to do is to offer 2 versions of the same architecture, one with HW, and the other with SW fl. pt. ops.
arithmetic operations on fl. pt. values compute results that cannot be represented in the given amount of precision. So, we round results. There are MANY ways of rounding. They each have "correct" uses, and exist for different reasons. The goal in a computation is to have the computer round such that the end result is as "correct" as possible. There are even arguments as to what is really correct. First, how do we get more digits (bits) than were in the representation? The IEEE standard requires the use of 3 extra bits of less significance than the 24 bits (of mantissa) implied in the single precision representation. mantissa format plus extra bits: 1.XXXXXXXXXXXXXXXXXXXXXXX 0 0 0 ^ ^ ^ ^ ^ | | | | | | | | | - sticky bit (s) | | | - round bit (r) | | - guard bit (g) | - 23 bit mantissa from a representation - hidden bit This is the format used internally (on many, not all processors) for a single precision floating point value. When a mantissa is to be shifted in order to align radix points, the bits that fall off the least significant end of the mantissa go into these extra bits (guard, round, and sticky bits). These bits can also be set by the normalization step in multiplication, and by extra bits of quotient (remainder?) in division. The guard and round bits are just 2 extra bits of precision that are used in calculations. The sticky bit is an indication of what is/could be in lesser significant bits that are not kept. If a value of 1 ever is shifted into the sticky bit position, that sticky bit remains a 1 ("sticks" at 1), despite further shifts. Example: mantissa from representation, 11000000000000000000100 must be shifted by 8 places (to align radix points) g r s Before first shift: 1.11000000000000000000100 0 0 0 After 1 shift: 0.11100000000000000000010 0 0 0 After 2 shifts: 0.01110000000000000000001 0 0 0 After 3 shifts: 0.00111000000000000000000 1 0 0 After 4 shifts: 0.00011100000000000000000 0 1 0 After 5 shifts: 0.00001110000000000000000 0 0 1 After 6 shifts: 0.00000111000000000000000 0 0 1 After 7 shifts: 0.00000011100000000000000 0 0 1 After 8 shifts: 0.00000001110000000000000 0 0 1 The IEEE standard for floating point arithmetic requires that the programmer be allowed to choose 1 of 4 methods for rounding: Method 1. round toward 0 (also called truncation) figure out how many bits (digits) are available. Take that many bits (digits) for the result and throw away the rest. This has the effect of making the value represented closer to 0.0 example in decimal: .7783 if 3 decimal places available, .778 if 2 decimal places available, .77 if 1 decimal place available, .7 examples in binary, where only 2 bits are available to the right of the radix point: (underlined value is the representation chosen) 1.1101 | 1.11 | 10.00 ------ 1.001 | 1.00 | 1.01 ----- -1.1101 | -10.00 | -1.11 ------ -1.001 | -1.01 | -1.00 ----- With results from floating point calculations that generate guard, round, and sticky bits, just leave them off. Note that this is VERY easy to implement! examples in the floating point format with guard, round and sticky bits: g r s 1.11000000000000000000100 0 0 0 1.11000000000000000000100 (mantissa used) 1.11000000000000000000110 1 1 0 1.11000000000000000000110 (mantissa used) 1.00000000000000000000111 0 1 1 1.00000000000000000000111 (mantissa used) Method 2. round toward positive infinity regardless of the value, round towards +infinity. example in decimal: 1.23 if 2 decimal places, 1.3 -2.86 if 2 decimal places, -2.8 examples in binary, where only 2 bits are available to the right of the radix point: 1.1101 | 1.11 | 10.00 ------ 1.001 | 1.00 | 1.01 ----- examples in the floating point format with guard, round and sticky bits: g r s 1.11000000000000000000100 0 0 0 1.11000000000000000000100 (mantissa used, exact representation) 1.11000000000000000000100 1 0 0 1.11000000000000000000101 (rounded "up") -1.11000000000000000000100 1 0 0 -1.11000000000000000000100 (rounded "up") 1.11000000000000000000001 0 1 0 1.11000000000000000000010 (rounded "up") 1.11000000000000000000001 0 0 1 1.11000000000000000000010 (rounded "up") Method 3. round toward negative infinity regardless of the value, round towards -infinity. example in decimal: 1.23 if 2 decimal places, 1.2 -2.86 if 2 decimal places, -2.9 examples in binary, where only 2 bits are available to the right of the radix point: 1.1101 | 1.11 | 10.00 ------ 1.001 | 1.00 | 1.01 ----- examples in the floating point format with guard, round and sticky bits: g r s 1.11000000000000000000100 0 0 0 1.11000000000000000000100 (mantissa used, exact representation) 1.11000000000000000000100 1 0 0 1.11000000000000000000100 (rounded "down") -1.11000000000000000000100 1 0 0 -1.11000000000000000000101 (rounded "down") Method 4. round to nearest use representation NEAREST to the desired value. This works fine in all but 1 case: where the desired value is exactly half way between the two possible representations. The half way case: 1000... to the right of the number of digits to be kept, then round toward nearest uses the representation that has zero as its least significant bit. Examples: 1.1111 (1/4 of the way between, one is nearest) | 1.11 | 10.00 ------ 1.1101 (1/4 of the way between, one is nearest) | 1.11 | 10.00 ------ 1.001 (the case of exactly halfway between) | 1.00 | 1.01 ----- -1.1101 (1/4 of the way between, one is nearest) | -10.00 | -1.11 ------ -1.001 (the case of exactly halfway between) | -1.01 | -1.00 ----- NOTE: this is a bit different than the "round to nearest" algorithm (for the "tie" case, .5) learned in elementary school for decimal numbers. examples in the floating point format with guard, round and sticky bits: g r s 1.11000000000000000000100 0 0 0 1.11000000000000000000100 (mantissa used, exact representation) 1.11000000000000000000000 1 1 0 1.11000000000000000000001 1.11000000000000000000000 0 1 0 1.11000000000000000000000 1.11000000000000000000000 1 1 1 1.11000000000000000000001 1.11000000000000000000000 0 0 1 1.11000000000000000000000 1.11000000000000000000000 1 0 0 (the "halfway" case) 1.11000000000000000000000 (lsb is a zero) 1.11000000000000000000001 1 0 0 (the "halfway" case) 1.11000000000000000000010 (lsb is a zero) A complete example of addition, using rounding. S E F 1 10000000 11000000000000000011111 + 1 10000010 11100000000000000001001 ------------------------------------------- First, align the radix points by shifting the top value's mantissa 2 places to the right (increasing the exponent by 2) S E mantissa (+h.b) g r s 1 10000000 1.11000000000000000011111 0 0 0 (before shifting) 1 10000001 0.11100000000000000001111 1 0 0 (after 1 shift) 1 10000010 0.01110000000000000000111 1 1 0 (after 2 shifts) Add mantissas 1.11100000000000000001001 0 0 0 + 0.01110000000000000000111 1 1 0 -------------------------------------- 10.01010000000000000010000 1 1 0 This must now be put back in the normalized form, E mantissa g r s 10000010 10.01010000000000000010000 1 1 0 (shift mantissa right by 1 place, causing the exponent to increase by 1) 10000011 1.00101000000000000001000 0 1 1 S E mantissa g r s 1 10000011 1.00101000000000000001000 0 1 1 Now, we round. If round toward zero, 1 10000011 1.00101000000000000001000 giving a representation of 1 10000011 00101000000000000001000 in hexadecimal: 1100 0001 1001 0100 0000 0000 0000 1000 0x c 1 9 4 0 0 0 8 If round toward +infinity, 1 10000011 1.00101000000000000001000 giving a representation of 1 10000011 00101000000000000001000 in hexadecimal: 1100 0001 1001 0100 0000 0000 0000 1000 0x c 1 9 4 0 0 0 8 If round toward -infinity, 1 10000011 1.00101000000000000001001 giving a representation of 1 10000011 00101000000000000001001 in hexadecimal: 1100 0001 1001 0100 0000 0000 0000 1001 0x c 1 9 4 0 0 0 9 If round to nearest, 1 10000011 1.00101000000000000001000 giving a representation of 1 10000011 00101000000000000001000 in hexadecimal: 1100 0001 1001 0100 0000 0000 0000 1000 0x c 1 9 4 0 0 0 8 A diagram to help with rounding when doing floating point multiplication. [ ] X [ ] -------------------- [ ] [ ][ ][ ] result in the g r combined to given produce precision a sticky bit
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