So today it is the last lecture in the series
of lectures on arithmetic operation and arithmetic unit design. Today we will be talking of floating point
operations. So, when numbers cannot be represented as
integers you represent them as floating point numbers. And we will see how one could do the usual
arithmetic and other comparison operations on floating point numbers. So we have been talking of primarily integer
arithmetic; we also talked about logical operation and other operations. I will first of all talk about why we need
to floating point numbers; what kind of representation we have and then we will work about the hardware
unit for carrying out the operation. So first what is need for floating point numbers;
how we represent them? I will talk about a standard representation
which is IEEE 754 then define the concept of overflow and underflow in case of floating
point numbers which is different from that we have in integers; how operations are carried
out; we will talk over primarily add, subtract, multiply and divide and talk little about
comparison as well. The issue of accuracy is very important in
floating point unit; how we can get best accuracy by using reasonable cost that will be discussed. There are some special numbers which need
special care we will talk of that and finally I will conclude by talking of what kind of
provisions are there in processor like MIPS. So first let us look at why we need to go
beyond the integers. We are familiar different data types as we
encounter in programming languages and also in the world of mathematics. We talk of integers which are actually a subset
of rational numbers, which is subset of real number and then subset of complex number so
it is a large space of numeric world and integers are only a small subset so how do you go beyond
integers that will be the topic of today. We need to also take care of situations when
the numbers are very large in terms of their values. One is that you may like to represent something
which is not quite an integer. Let us say 7.4 which is neither 7 nor 4 but
on the other hand, there is also a need for representing values which are extremely large
or extremely small. Particularly in scientific domain we have
to talk of huge distances. for example, in astronomy if you are looking at distance between
let us say Pluto and Sun this is 5.9 into 10 raised to the power 12 meters so you would
not like to represent that as an integer; it is not that you are interested in values;
even if your accuracy requirement is limited you are not looking at fractional values but
the number is so large that integer representation is not the suitable one. On the other extreme if you look at quantity
like mass of electron which is 9.1 into 10 raised to the power minus 28 grams again it
is a so small value that you will either have to scale it to that extent or think of a different
representation. So how do we represent fractions; that is
one question and how do we accommodate large range that is the second part. Representing fraction means you want something
which has an integer part or there is a part which is a non-integer. Also, there could be you may be looking at
non-integer values as rational number. For example, you have rational numbers they
can be thought of integer pairs so 5 or 8 can be represented by two integers. Or if you have a number like let us say minus
247.09 you can represent this as a string. Minus is one character, 2 is one character,
4 is another, 7 another and so on. So each of these boxes which you see is a
separate character so as a string; you can represent a number which has something to
the left of the decimal and something to the right of the decimal and an appropriate sign. Or in binary form you can represent a number
with a fixed position of the binary point. For example, look at this number string of
1s and 0s and you might say that here is my binary point which is equivalent to.... which
is the counterpart of decimal point. So you may not, unlike this representation
where you are using a character to represent that position this position may be implicit
so you would say that out of so many bits the point actually lies after the fifth bit
from the right then you need to only worry about the string of so many bits and the position
of the binary point is fixed so this could be consider as fixed point notation. But this representation although is okay it
does not take care of large range. So, to get a large range both in terms of
extremely large value and extremely small value you need what is called floating point
where you have a fraction part which could be represented in any of the manners described
above and there is a base to which you can specify you can raise to a specified power. So it is this ability that you have a base
and some power associated with it that you can work with large ranges. You can have a positive power to represent
extremely high values and negative power to represent extremely low values. The precision or the accuracy of what you
are representing would actually come from this fraction part; how many significant digits
you have would come from this whereas this power will take care of the range. Now we must remember that in all such representation
where you are using a finite number of bits or digits you are representing basically values
which are rational. To represent irrational quantities you need
infinite length of storage. So, as far as irrational numbers are concerned
we would typically represent only a reasonable approximation of those numbers and all representation
which are infinite number bits or bytes would be actually representing rational number in
some sense. So now let us a go deeper and understand the
meaning of binary numbers where there is a point somewhere. So, for example, take a simple number 101.11
so there is a this is the integer part and there is a fractional part. What is the value of such a number? We understand very well the part two the left
of the point 1 into 2 raised to the power 2; 0 into 2 raised to the power 1 and 1 into
2 raised to the power 0 corresponding to this 1 0 and 1. The bits on the right of the point have weightage
which is negative powers of 2. So this 1 here has a weightage of 2 raised
to the power minus 1 and the next one has 2 raised to the power minus 2. So, if you add now all these bits with appropriate
weightage which be which is a power of 2 positive or negative you get this 4 that this is the
4 0 that gives you 1 that gives you 0.5 that gives you 0.25 and all put together we get
5.75 in base 10 so this shows how you can actually look at binary fraction like this
and get its equivalent decimal value. On the other hand, suppose you have a decimal
fraction see 0.6 how do you get its binary representation? Here is the representation but let us go through
the steps which will give you this. The process is that you repeatedly multiply
this fraction and keep on collecting keep on multiplying by 2 and keep on collecting
the integer parts. You multiply 0.6 by 2 you get 1.2 so 1 is
the integer part and 0.2 is the remaining fractional part. then 0.2 is again multiplied by 2 you get
0.4 so look ahead is at 0.4 so 0 is the integer part 0.4 is the fractional part; repeat the
process you get 0.8, 1.6 and now 0.6 since you got 0.6 it will again follow same steps. So, in terms of it is binary representation
you will get 0.1001 and this will repeat; just to show the repetition I have shown different
colors. So now this means that you repeat this still
infinity. But if you are talking of finite number of
bits again this cannot be accurately represented because you will have to truncate it somewhere. Therefore, we cannot represent 0.6 exactly
in binary by this method. Of course one of things which I mentioned
in the previous slide is that, rational number you will represent as a pair of integers that
could be done. You could represent this as 6 and 10 that
would be an exact representation but that is not very convenient to work with. What is convenient to work with is something
like this but there is some loss of accuracy which we must keep in mind. Now let us introduce the exponent part. You have some base and you raise it to some
power and also let us bring in the sign. So typically, a floating point representation
would have a form like this so minus 1 raised to the power s, s takes care of the sign s
is either 0 or 1 multiplied by F, F is the fractional part so fraction is in some fixed
point sum that means you have a string of bits and there is some position where you
assume binary point to be there. The base here is 2; we are talking a binary
now so 2 raised to the power E, E is the exponent part, this F is the fractional part which
is usually called mantissa or significand and E is the exponent part; this itself can
be positive or negative we have seen in need for both why we need positive exponent; why
we need negative exponent. So now with this concept the question is that
a given word of memory or a given register how do you divide this into three parts. That means how you allocate bits to S, F and
E, S of course is straightforward it requires only one bit but how many bits for S, how
many bits for E so there could be lots of options; lots of options are there and also
there could be lots of options as how you represent F where do you assume the point
to be and how do you represent E what about the sign of E so there are all these issues
and there are multiple answers. So what has happened is that over a period
of time a standard has emerged which is largely acceptable and today most of the processors
actually use that standard. This is IEEE 754 standard. IEEE stands for Institution of Electrical
and Electronics Engineers. This is an organization, which, apart from
publishing the scientific literature, they also define various standards. So this standard defines two floating representation
which is called single precision and double precision. Double precision naturally has larger number
of significant digits. So the division of the word which is 32-bit
is as follows. The left most bit is the sign bit, next 8
bits are kept for exponent and the remaining 23 bits are for F or the mantissa. In double precision we not only extend the
precision but also increase the range of E. So there are three additional bits given here. This small difference you will see makes a
big impact so you do not need to really increase too much here. so 11 bits for E and the remaining bits of
this word and another word; it is a total 64-bit representation for double precision
so 20 remaining bits here and 32 of the next word so together 52 bits are used for the
F part or the mantissa part. Now let us go into further details of how
E and F are represented. So first let us look at F. F is the significant
part or the mantissa part where in single precision case we have actually 23 bits; we
assume the point to be on the extreme left and we also assume that there is a one sitting
there invisible. There is an implicit 1 here which means that
the numbers here we are going to talk of will be 1.something so this would decide the range
as you would see that smallest number would be when it is 1. All 0s and the largest value of F would be
when it is 1.followed by all 1s; we will see what it means. Similarly, in double precision again the point
is to the left extreme and there is an invisible 1 here; invisible or implicit that means you
assume that 1 to be there. So effectively we are having 24 bits here;
1 bit we do not have to represent explicitly. So now what are the value ranges? Naturally with all 23-bit 0 the value of F
will be 1 1.00 or effectively 1. And if you have, get in back to this if you
have all the 1s here then what is the value; it is nearly 2 but not quite equal to 2; how
much short of 2 it is? Corresponding to a 1 in this position. So, if you take 1.all 1s and then add a 1
in this position you will have carries going all through and you will get a 2 here. We are only slightly short of 2 and to be
more precise it is 2 minus 2 raised to the power minus 23 because that is the weightage
of the last bit twenty third bit it is equivalent to 2 raised to the power minus 23. So you could say that F lies between 1 and
2 minus 2 raised to the power minus 23 or more approximately you could say that 1 is
less than equal to F less than 2 so we are not saying how less but we know that it is
nearly almost 2. Similarly, for double precision numbers except
for 23 in place of that we put 52. So now this way of representing the floating
point number or the mantissa of this is called normalized representation. So, if the value actually falls beyond this;
suppose you are talking of you want to say 5 into 2 raised to the power something so
instead of saying 5 you divide by suitable power of 2 so that the value gets in the range
of 1 and 2 so it will be actually 1.25 and you make the appropriate adjustment in the
exponent. So, suppose you wanted to say 5 into 2 raised
to the power 4 equivalently you could say 1.25 into 2 raised to the power 6 so you divide
here by 2 raised to the power 2 and adjust it by increasing the exponent. Similarly if F happens to be very small you
multiply it by powers of 2 suitably and decrement the exponent appropriately so that the value
always lies in the range 1 and 2; the value of F is always in between 1 and 2. If it is exactly 2 you can divide by 2 and
make it 1 and adjust the exponent. Let us now look at E part or the exponent
part. For single precision numbers we have 8 bits
here. We represent exponents with a biased notation. The signed exponents are not represented as
2's compliment there is a reason I will come to I will explain little later. So the bias used here is 127. That means if your exponent is minus 126 minus
126 plus 127 it will appear as 1. So the values these 8 bits actually can carry
a value from 0 to 255 so leaving the two extremes 0 and 255 which are for special purpose I
will explain that later; the real range is 1 to 254 and 1 corresponds to the most negative
exponent, 254 corresponds to most positive exponents and with the bias of 127 1 actually
corresponds to minus 156 and 254 corresponds to plus 127 Similarly, in double precision numbers the
bias is 1023 and the range will go from minus 102 to plus 1023 that is shown here. The range in range of E in single precision
is minus 126 to 127 we are excluding the cases when all E bits are 0 or all E bits are 1
that is the two extremes are used for something special and we will discuss that later. Now we have talked of positive and negative
numbers what about 0? So 0 would require actually all F bits to
be 0 and we cannot assume an explicit 1 to the left of the decimal to left of the point
because that would mean it is 1.something; that range excludes 0 so 0 again has to be
treated specially. So what we say is when F is all bits 0 and
E is also all bits 0, signed bit is also 0 then it actually represents a 0 in floating
point. So floating point 0 is also same as integer
0 in representation. All 32 bits are 0 it signifies floating point
0. In principle any number which has f0 and any
value of exponent would represent 0 but that would correspond to multiple representations. So we have a unique representation in IEEE
754 which is all bits are 0. So now with this we described how do we test
the numbers for being zero, non-zero, positive, negative, how do we compare and so on. So test for 0 as you can now guess is same
as what you do for integer. So, if you have a circuit which tests a number
for integer 0 it will naturally test for floating point 0 also. Similarly, tests for sign test for being negative
or positive requires only the sign bit to be look at which is same in the two cases:
integer as well as floating point. What about a magnitude comparison? Now, magnitude comparison also turns of to
be similar to integer because of certain choices we have made. Firstly we have kept exponent part to the
left of the mantissa part. So now, if two numbers have same mantissa
part but one has a larger exponent obviously that will be larger because exponent has a
larger weightage. Even if two numbers have different exponents
sorry different mantissas let us say A has larger mantissa than B but if B has a larger
exponent B will be larger. So it is the exponent which is more predominant
and in terms of significance it naturally places towards the left. So therefore, when you compare to floating
point numbers treating them as integers and if you find that exponent bits in one are
larger than the exponent bits in the other then you do not need to really look at the
mantissa so that is how it will happen in case of integer. Also, if some MSBs give you a decision about
A in larger than B or B in larger than A then you do not need to look further. And another thing which helps in this is that
the bias representation. So, irrespective of sign of the exponent this
comparison will work out. If you have, for example, 2's compliment representation
then the signed comparison and unsigned comparison differ. So now we can do magnitude comparison of two
floating point numbers which means we are not we are excluding the sign bit rest we
compare as if we were comparing magnitudes of integers. In integers we talked about overflow; we talked
of overflow when you are trying to get to a very large positive integer or very large
negative integer so if we exceed the positive limit or negative limit you have overflow. Here we have additional concept of underflow. So, if we look at the largest and the smallest
positive or negative numbers we can represent you would notice that there is a limit of
values in both directions. So you may have a number which is which is
larger than the largest positive number, for example, so that is an overflow condition. But if you have a number which is not 0 but
smaller than the smallest floating point number you can have; it is too small to be represented
in the in the given floating point notation then it is underflow. So, number is not 0 but it cannot be. In integer the smallest integer is 1 and you
can represent that so there is no underflow but here there is a concept of underflow which
you must remember. So now let us look at the extreme values. We take the largest value of the mantissa
which was 2 into 2 minus 2 raised to the power minus 23 or approximately 2 and largest positive
power is 127. So, basically, it is 2 into 2 raised to the
power 127 that is the largest number we have and 2 raised to the power 127 turns out to
be approximately 10 raised to the power 38. So in terms of familiar decimal system that
is the range we are getting; we are we can go up to power 38 of 10. The smallest positive or negative integer......
so here as far as the overall sign is concerned that is the separate bit so I am not writing
separate range of positive and negative; I am just prefixing a plus minus sign with this. So the smallest positive and negative value
is we take the smallest F value which is 1 and the smallest E value which is minus 126
so this can also be written as 2 into 2 raised to the power minus 127 which means it is 2
raised to the power 2 into 10 raised to the power minus 38. So basically that is the range powers of 10
going from minus 38 to plus 38. In the double precision here it is a power
2 2 raised to the power 1023 which turns out to be 10 raised to the power 308. So it is extremely a large number which is
good enough for almost all practical purposes. Another thing you need to be concerned here
is how many significant digits are there, let me see if I have................. another
question is how many significant digits you have. See, we have significant bits as 23 here and
52 here so what does it translate to equivalent decimal digits. So to get an idea of the precision of the
number you are trying to represent you can actually divide this 23 by log of 10 to the
base 2 and you will get the number of digits. Similarly divide 52 by log of 10 to the base
2 and you will get the number of significant digits. So I think this will be something like 8 digits,
7 or 8 digits or so on. Now finally let us come to how we perform
operations on floating point numbers. So first let us look at addition and subtraction. Let one number be minus 1 raised to the power
S1 multiplied by F1 multiplied by 2 raised to the power E1. So S1 F1 E1 are the three components of this
floating point number. Similarly, there is the other one S2 F2 and
E2 and we are required to either add or subtract. Before we can do this addition subtraction
we must bring the exponents to the same level. Suppose E1 is greater than E2 find out which
is larger let E1 be greater than E2 then we can rewrite the second number in such a manner
that its exponent is made even; we do over with normalization here, change the fraction
part to F2 prime where F2 prime is obtained by dividing F2 with some power of 2. So, specifically 2 raised to the power.....
the difference is of the two exponents. So now dividing by power of 2 actually you
can imagine that in hardware circuit it would mean that you are shifting right. So you shift F2 by so many positions E1 minus
E2 and corresponding adjustment has been done in the exponent; exponent has gone from E2
to E1. So now you are in a position two adder subtract
F1 and F2 prime. So, depending upon what the signs are and
what the operation is you will be performing one of these operations: sum or a difference
so the result is minus 1 raised to the power S1 the sum or difference of these into 2 raised
to the power E1. Well, by the way this itself could be positive
or negative and accordingly that might also change. So the logic for determining sign is not difficult
to work out. What I like to point out here is that when
you do this you may find that this is not normalized; this sum or difference might go
beyond the range of 1 to 2 which we would like to have. So, therefore, this operation has to be followed
by normalization process. If the number is become too large shift it
right and adjust the exponent; if it has become too small shift it left and adjust the exponent
in the opposite manner. So here this is not a detailed a circuit but
a kind of bock diagram indicating the flow of information as it would be in a floating
point adder cum subtractor. So let us say we have these two registers
holding the two numbers sign exponent and significant part. The first thing which will be done is compare
the exponent; the circuit comparing these is called small ALU because it is looking
at only 8-bit operands. It computes the difference of the two exponents
and that is the amount by which you have to shift one of the significands so that the
exponent gets aligned. This is called the alignment stage. Here there is a shifter which will perform
right shift on one of the significands and that is selected by this multiplexer and which
one selected is determined by this control unit and the control unit is guided by the
difference which you have found out. So knowing whether E1 is greater than E2 or
E2 is greater than E1 one of the two will be selected here, other one will be selected
to go directly and this big ALU adds or subtracts the two significands or mantissas and you
get the difference here. so this is This you could say is the first
stage where the exponents are getting compared; this is the second stage where the number
with smaller exponent is brought in par with the one with a large exponent. This is the stage where this is the stage
where addition or subtraction is done, this is the stage where normalization takes place. So normalization would involve output of ALU......
ignore these multiplexers for the moment I will come back to this in a moment; this output
to the ALU may need to be shifted left or right depending upon how far you are from
the normalized value so you need to bring it between the range of 1 and . And accordingly
there will be an increment or decrement of the exponent which will be done which is actually
coming from here; the one of the two exponents selected and passed out for increment or decrement. Now, after normalization you need to what
is called rounding. Rounding means you like to get to the LSB
to the best of the accuracy. So you may be throwing something towards the
right and whether you need to make an adjustment here or you can just afford to through it. I will come to this shortly. You have some rounding logic here which may
further adjust the mantissa; it may need adjustment in the exponents and after rounding off you
may find that we have lost normalization once again so there may be another round of normalization
so that is why these are being fed back and the multiplexer will select either the original
value coming from top or after rounding and finally the result can be placed in another
register. This is the hardware for add or subtract. Conceptually multiply operation is somewhat
simpler. If you look at two numbers: this is one, this
is second, their product can be obtained straightaway, the sign of the result is Exclusive OR of
the signs, if both are 1s or both are 0s the result is 0 and 1 otherwise. The fraction part is the product of the two
fractions and the exponents gets summed so it rather straightforward there is no initial
alignment which is required. But of course you may need to do normalization
and rounding the reason for that is this product of two values which are individually in the
range 1 to 2 the product could be in the range 1 to 4
so that means on the larger side it can exceed 2 and it may require one smaller adjustment
you may need to bring it down. If it has exceeded 2 you divide it by 2 and
increment the exponent terms. Divide is similar; two numbers same thing,
the sign is determined exactly in the same way, the fractions get divided, the exponents
get subtracted. Now this ratio of the fractions could lie
in the range 0.5 to 2 so it will not exceed 2 because each one is in the range 1 to 2
but it can become small so you would you may need to normalize it here and again there
is a there is always a need for rounding off. So of course before you do all that you need
to check whether F2 is 0 or not. So you can proceed with this if F2 is not
0. Now the question of accuracy and issue of
rounding off.......so when when while doing the operations, for example, when you are
aligning you are shifting the number to the right, you are throwing of some bits. When you are multiplying two fractions you
have two 32-bit or let us say two 24-bits including the invisible 1 bit. You have two 24-bit exponents sorry 22 24-bit
fractions which you are multiplying the result would be, as we have seen for integers, the
result would be 48-bits in general. So you have to now throw the remaining bits
and retain the 24-bits finally. So there is a loss of precision in this process
because you are working with finite word length and of course you have to have finite word
length so some loss would be there but you want to minimize the loss. In the given number of bits what is the best
you can do. So it turns out that if you use a few extra
bits for intermediate calculation and finally do a round off based on those then you can
get to the accuracy which is theoretically possible. So these bits are called G, R or S. G stands
for guard bit, R for round bit and S for sticky bit. So G and R are nothing but the next two bits
after those 24 bits. We normally have 24 bits as seen externally
but suppose we had if we had retained two more bits these are G and R. So, in the intermediate
calculation we retain those. Apart from this we keep one more bit which
is called S bit and S bit is 1 if and only if any bit towards right of R is non-zero. So S distinguishes; S is like a summary of
the remaining bits which we are throwing off and tells it distinguishes the case which
is all zeros and case where there is some at least a single 1. You can look upon the number as 1.all these
followed by three more bits G, R and S. Now, given these bits; given these additional
bits which are which are kept in the intermediate stages how do you perform rounding off. So now first let us look at G and R and just
for relative sense let us imagine that the point is to the left of G. So, on a relative level let us assume that
the binary point is here actually; it is just for concept otherwise actually the point is
there. So G has a weightage of in relation to this
G has a weightage of 0.5, R has a weightage of 0.25. So when G and R both are one it is like 0.75
and you will round up you will add 1 to the LSB. Now LSB means the bit to the left of G. if
both are if G is 0 and R is 0 it is the case when both are 0, I mean it represents 0 to
the right of the point or when R is 1 but G is 0 it corresponds to 0.25 so when it is
0.25 or 0.00 now you want to throw away the value towards the right of the point you want
to round it down. When G is 1 and R is 0 it corresponds to 0. 5 you are exactly in the middle so whether
you go up or you go down; in this case we will look at S-bit so we want to get more
information to be fair. If S is 1 that means there is something which
is non-zero towards the right of R so we can round up we can add 1 to LSB; if S is also
0 then we are dead at 0.5 and which way do we go? So now assuming that half the time you want
to round up half the time you want to round down and all numbers are equal likely the
logic which is followed in IEEE 754 is that you round to the nearest even; even in the
sense of this bit this bit being considered as LSB so 1 here means it is odd and 0 here
means it is even. So if it is 0 here you will you leave the
number as it is, if it is 1 here it is like odd so you add 1 to make it even. So this is the meaning of round to the nearest
even. So you round in a manner that the last bit
twenty third bit becomes 0. So effectively what you are doing is you are
trying to be fair. Now such things are actually important, for
example, when you are dealing with the financial applications so if you if you if there is
a bias in your rounding method then over a period of time the gains may accumulate for
one party and losses may accumulate for another party which may not be fair so you would like
that rounding up takes place half the time rounding down takes place the other half of
the time. I mentioned that some codes for exponents
are reserved for special numbers so here you will see the entire picture. The first two columns are for single precision
numbers and the next two columns are for double precision number and the last one represents
last one indicates what object or what kind of number we are trying to represent. So if you have everything as 0 when it is
a special case it is representing a 0, when...... let me look at this..... when exponent is
at it is peak high the highest value 255 and mantissa is 0 then it represents infinity. The sign the sign bit is also there which
could be plus or minus and it will represent accordingly plus infinity or minus infinity
so now what do programs do with infinity. When you are writing a program you can always
check this condition. For example, at some point you let us say
reach a situation that the result is coming as infinity how do you represent; you do not
want to stop computation there, you want to represent infinity and proceed further. So you can write you can express this in this
way and subsequent part in the program would understand that the number is infinity so
you will not try adding infinity to something which is not infinity and so on so this gives
a method for doing so. The other combination, for example, suppose
exponent is 0 but mantissa is non-zero this represents a denormalized number. So when underflow occurs you can still proceed
with your computations; you can represent numbers by doing away with the requirement
of normalization because if you insist in normalization then there is a smallest value
you cannot go beyond that so you do go beyond you force your way beyond that represent them
as denormalized numbers but you must know that now there is that invisible 1 is not
there. This is the normal situation that exponent
is the range 1 to 254 and mantissa could be anything and the last one is that exponent
is 255 same as infinity but you have a non-zero value. So this is the this is used to represent something
which is not a number. For example, 0 by 0 you cannot say it is infinity
it is something which is undefined. So there could be different situations which
lead to numbers which are undefined and by choosing some code here you can have programs
and specific meaning of all those situations. Finally I will give a summary of the instructions
which are there in the MIPS processor to handle floating point number. First of all there are thirty two floating
point registers labeled as dollar f0 to dollar f31. These are in addition to the thirty two register
we have for integer. Actually this entire hardware which does floating
point activity is considered as a co-processor so it is a companion of the processor whose
special task is to work with floating point numbers. So there are usual operations: add, subtract,
multiply, divide, absolute value, negate which can work on single precision data or multi-precision
double precision data. So you have two versions of each add dot s
and add dot d stands for single precision addition or double precision addition. So, like integer MIPS addition this requires
three floating point registers to be specified. You would say add dot s and then give three
registers. When you are working with double precision
arithmetic you need to specify registered pairs; f0 f1 form one pair; f2 f3 form another
pair and so on there are sixteen pairs; then there are instructions lwc1 and swc1 so this
is load word but we are also saying c1 which stands for co-processor 1. There can be many co-processors attached with
the main processor and floating point processor is considered as co-processor 1. So these instructions would load a word from
memory into the co-processor 1 which is the floating point unit that means one of these
registers so you specify which registers you want to load into and you specify one integer
register in the usual manner to specify the address so there is a constant part 16-bit
constant part and a base register so together they specify memory address; the store instruction
is similar. Then there are conditional branches bc1t and
bc1f. b stands for branch, c1 stands for co-processor
1 and t and f stands for true and false condition. Now how do you determine the condition? You need to use one of these compare instructions
to set the condition. Imagine that condition is a 1-bit register
or a flag which is set by one of the compare instructions. So, in SLT for example, in SLT instructions
the result of comparison was brought into one of those thirty two registers but here
the result of comparison is brought into a separate flag which is tested by this so test
for truth and test for falsehood. These two instructions do the comparison so
c means compare, then lt means less than, s or d would mean single precision or double
precision. So lt could be replaced by le gt ge so there
are six different instructions for single precision, six for double precision. Then there are instructions for conversions. Because you have three representations: integer,
single precision and double precision and you need conversions. Suppose you have a number 3.5 it will have
some representation in single precision, some representation in double precision so a 32-bit
pattern or a 64-bit pattern and there may be sometimes a need for conversions or conversion
to integer which means to take out the integer part of if you have integer 3 you can convert
to floating 3.0 so these conversions could be somewhat lossy; particularly when you are
going from higher precision to lower precision you may lose some information. Finally what we have learnt today is we talked
about range of values which we can handle, definition of overflow and underflow, we learnt
how to perform various operations, we talked about accuracy use of G, R and S bit is for
rounding off, we also looked at notations for some special numbers and finally looked
at floating point operations given by different instructions in MIPS processor. Thank you.
