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MIPS R2000 system consists of a CPU, and a separate floating-point unit (FPU) referred as Coprocessor 1. Coprocessor 0 handle exceptions and other tasks.

Floating-point registers reside in FPU. There are 16 floating-point registers on MIPS used for floating-point calculations. They are called $f0, $f2, $f4, ..., f$30. Only even numbered registers should appear in most of the instructions. If the number is a single precision number, the even numbered register is used. Double precision number use two floating-point registers, e.g., $f2, $f3.

single precision double precision add add.s add.d subtract sub.s sub.d multiply mul.s mul.d divide div.s div.d load l.s l.d store s.s s.d load immediate li.s li.d compare lt c.lt.s c.lt.d eq, le, ... c.eq.s c.eq.d branch on comparison true bc1t branch on comparison false bc1f move from FP reg to general reg mfc1 move to FP reg from general reg mtc1Please refer to Fig. 4.44 and Appendix A of the textbook for syntax and complete list.

Write an assembly procedure poly to evaluate the value of the polynomial

f = a[0] + a[1] * x + a[2] * x^2 + ... + a[n] * x^n.where a[i] is given as an array of float type coefficients, n is an integer for the order of the polynomial, and x is the variable of type float.

An efficient method is the Horner's method, by noting

f = a[0] + a[1] * x + a[2] * x^2 + ... + a[n] * x^n = a[0] + x * ( a[1] + x * ( a[2] + ... + x * ( a[n-1] + x * (a[n] + 0))...)we can compute with the following steps (algorithm)

i = n; y = 0; Loop: y = a[i] + (x * y); i = i - 1; if (i>=0) goto Loop; return y.

With the Horner algorithm for polynomial evaluation given above, we can write the assembly program as below. We'll write the program as a function (or called procedure). We assume that the address of a is in $4, the value n in $5, x is loaded in $6. The result is put in $f0. The register usage for passing values into a function or return result of the function is not arbitrary, but follows certain convention. We'll discuss this later in Appendix A.

poly: li.s $f0, 0.0 # y = 0, running and return result mtc1 $6 $f12 # x, move to float register Loop: mul.s $f14, $f12, $f0 # compute (x * y) mul $2, $5, 4 # $5 = i, compute address of a[i] addu $3, $2, $4 # a + (i*4) l.s $f16, 0($3) # a[i], load coefficient add.s $f0, $f16, $f14 # y = a[i] + (x*y) addi $5, $5, -1 # decrease i slt $2, $5, $0 # $2 = 1 if i < 0 beq $2, $0, Loop # goto Loop if i >= 0 j $31

(1) With integer comparison and branch, we use something like

slt $3, $4, $5 # $3 = 1 if $4<$5 bne $3, $0, L # go to L if condition true

(2) Floating-point comparison is treated differently, e.g.,

c.lt.s $f0, $f2 # set FPU condition flag to true if $f0 < $f2 bc1t L # branch if FPU condition flag is true.

Example: Add 2.56 x 10^0 to 2.34 x 10^2, assuming that we have three significant decimal digits, with and without guard and round digits.

With extra 2 digits (guard and round)

2.56 x 10^0 = 0.0256 x 10^2 + 2.3400 x 10^2 ------------------ 2.3656 x 10^2After rounding, the result is 2.37 x 10^2.

Without extra bits

2.56 x 10^0 = 0.02 x 10^2 + 2.34 x 10^2 ------------------ 2.36 x 10^2So, the answer is off by 1 in the last digit.

Method examples round to 3 significant figures Round up 300.45 301 (toward +infinity) -300.45 300 0.07622 0.0763 5.5550 5.56 Round down 300.45 300 (toward -infinity) -300.45 -301 0.07622 0.0762 5.5550 5.55 Truncate 300.45 300 (toward zero) -300.45 -300 0.07622 0.0762 5.5550 5.55 Round to nearest even 300.45 300 -300.45 -300 0.07622 0.0762 5.5550 5.56 5.5450 5.54The last rounding method is implemented in most computers. Round to the nearest picks the number nearest to the original number. When two numbers are equally near (e.g. 5.556 and 5.554 are equally near to 5.555), round (add 1 to the last digit) or truncate the number so that the result digit is even.

Floating-point numbers are approximation to the real numbers in mathematics. In single precision (float), 23 bits are used for the fractional part. Together with the implicit leading 1, we have 24 significant binary digits. Converting into decimal, it means that we have 24 * log_10 2 = 0.3 * 24 = 7 decimal digits. In other words, single precision is accurate to seven significant decimal digits. For double precision, it is precise with 53 * 0.3 = 16 significant decimal figures.

Since the computer representation of real numbers contains discrete and finite number of values, while real numbers are continuous and uncountably infinite many, most real numbers are represented only approximately. The maximum error is half the size of the gap between representable numbers.

Given a representable number x, the number just slight smaller (or larger) is the one that the significand f differs by 2^{-23} (last bit). So the gap size around x is

Absolute Error = gap/2 = 2^{e-127} delta f / 2 = 2^{e-127} 2^{-24} = x 2^{-24} = 6.0 10^{-8} x, Relative Error = Absolute Error / x = 10^{-7}.Note that the gap is proportional to x, and relative error is a constant 0.0000001, that is, there are seven significant figures.

Example: Given the number

x = 0 10110011 10000100101100001110110What are the numbers just slightly smaller and larger that x?

Answer:

x - delta = 0 10110011 10000100101100001110101 x + delta = 0 10110011 10000100101100001110111So the gap can be computed exactly as

delta = 2^{e-127} (0.00000000000000000000001_two) = 2^{e-127} 2^{-23}And the relative error (more precisely, maximum relative error) is

delta/(2*x) = 2^{-24} = 4 x 10^{-8}

**Problem**: (1) What is the decimal value of this
IEEE single-precision floating-point number?

0011 1111 0101 0000 0000 0000 0000 0000

Answer: s = 0, e = 01111110 = 126, f = 0.101 = 1/2 + 1/8 = 0.625. So the value is (1+f) 2^(e-127) = 1.625 x 2^(-1) = 0.8125.

(2) What is the IEEE single-precision floating-point representation for the decimal number -3.25.

Answer: 3 = 11_two, 0.25 = 0.01_two. So, 3.25 = 11.01_two = 1.101 x 2^1. Thus s = 1, e - 127 = 1, or e = 128, f = 0.101, the pattern is

1 10000000 10100000000000000000000

Besides the normal numbers, IEEE representation also has other types of symbolic entities to aid the computation. They are the infinities and NaN (not-a-number).

+infinity 0 11111111 00000000000000000000000 -infinity 1 11111111 00000000000000000000000

Infinities are represented by e=255, and f = 0. The infinities behave the way we expected. For example 1.0/0.0 = +Infinity, +Infinity - 10 = +Infinity. 1/Infinity = 0.

If e=255 (all ones), and f != 0, then the pattern is considered not a number! NaN can be generated during computation, e.g., 0.0/0.0 = NaN, Infinity - Infinity = NaN. If your program output produces a NaN, it is a good indication that something is wrong with your program.

There are two versions of zeros. Zeros are presented by

0 00000000 00000000000000000000000 (0+) or 1 00000000 00000000000000000000000 (0-)

(-1)^s x f x 2^{-126}This type of number is called denormalized since the leading digit is not 1.

Question: What is the smallest denormalized number in simple precision?

Answer: Smallest f is 2^{-23}, so the smallest denormalized number is 2^{-23} x 2^{-126} = 2^{-149} = 10^{-45}.

sign s takes 1 bit, exponent e takes 8 bits, and significand (fraction) takes the rest of 23 bits. According to the value of e, we have the following situations.

- Normalize number: e = 1 to 254. The value is (-1)^s x ( 1 + f ) x 2^{e-127}.
- Infinity and NaN: e = 255. If f = 0, value is (-1)^s * Infinity. If f != 0, the pattern is Not-A-Number.
- Denormalized number: e = 0, the value is (-1)^s x f x 2^{-126}.
- Zero is s = 0 or 1, e = 0, f = 0.

Pitfall: Forgetting that floating-point addition is not associative [hence x + (y + z) != (x + y) + z].

Example: x = 1.0, y = -1.0, z = 1.0E-8 = 0.00000001, in single precision. Then

x + ( y + z ) = 1.0 + (-1.0 + 1.0E-8) = 1.0 + (-1.0) = 0.0, (x + y) + z = (1.0 + (-1.0)) + 1.0E-8 = 0 + 1.0E-8 = 1.0E-8.

- Computer arithmetic has limited range and precision.
- Bit patterns have no inherent meaning.
- IEEE 754 floating-point standard.
- Floating-point instructions.

The textbook, "Computer Organization & Design", pages 238-257.

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