Chapter 3 Arithmetic for Computers

82 Pages·2009·5.17 MB·English

Chapter 3
Arithmetic for Computers
授課教師: 張傳育 博士 (Chuan-Yu Chang Ph.D.)
E-mail: [email protected]
Tel: [phone] ext. 4337
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IInnttrroodduuccttiioonn
自然數字的二進位表示法
二進位與十進位的轉換(binary to decimal conversion)
二進位與十六進位的轉換(binary to hexadecimal
conversion)
十六進位與十進位的轉換
負數表示法
二進位數的邏輯運算 (and, or)
二進位數的算術運算(+, -, *, /)
分數(fraction)及浮點表示法
硬體如何實際執行加、減、乘、除。
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Signed and Unsigned Numbers
In any number base, the value of ith digit d is
d × Base i
Binary numbers (base 2)
least significant bit: the rightmost bit
most significant bit: the leftmost bit
0000 0001 0010 0011 0100 0101 0110 0111 1000 1001...
decimal: 0...2n-1
Of course it gets more complicated:
numbers are finite (overflow)
fractions and real numbers
negative numbers
e.g., no MIPS subi instruction; addi can add a negative number
How do we represent negative numbers?
i.e., which bit patterns will represent which numbers?
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What is the expansion in storage if the number 1 billion
is represented in ASCII versus a 32 bit integer?
1 billion = 1000000000 (10個ASCII符號)
1 ASCII = 8 bits
所以需要8*10=80 bits
因此80/32=2.5
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Possible Representations of
Negative Numbers
Sign and Magnitude: One's Complement Two's Complement
000 = +0 000 = +0 000 = +0
001 = +1 001 = +1 001 = +1
010 = +2 010 = +2 010 = +2
011 = +3 011 = +3 011 = +3
100 = -0 100 = -3 100 = -4
101 = -1 101 = -2 101 = -3
110 = -2 110 = -1 110 = -2
111 = -3 111 = -0 111 = -1
Issues: number of zeros, ease of operations
Which one is best? Why?
The drawbacks of the sign-magnitude representation:
It’s not obvious where to put the sign bit.
Adders for sign and magnitude may need an extra step to set the
sign because we can’t know in advance what the proper sign will be.
Two zeros, positive and negative zero.
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MIPS: 2’’’’s complement
以0為前導的數為正值
32 bit signed numbers:
0000 0000 0000 0000 0000 0000 0000 0000 = 0
two ten
0000 0000 0000 0000 0000 0000 0000 0001 = + 1
two ten
0000 0000 0000 0000 0000 0000 0000 0010 = + 2
two ten
...
maxint
0111 1111 1111 1111 1111 1111 1111 1110 = + 2,147,483,646
two ten
0111 1111 1111 1111 1111 1111 1111 1111 = + 2,147,483,647
two ten
1000 0000 0000 0000 0000 0000 0000 0000 = – 2,147,483,648
two ten
minint
1000 0000 0000 0000 0000 0000 0000 0001 = – 2,147,483,647
two ten
1000 0000 0000 0000 0000 0000 0000 0010 = – 2,147,483,646
two ten
...
1111 1111 1111 1111 1111 1111 1111 1101 = – 3
two ten
1111 1111 1111 1111 1111 1111 1111 1110 = – 2
two ten
1111 1111 1111 1111 1111 1111 1111 1111 = – 1
two ten
以1為前導的數為負值
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ccoonnvveerrssiioonn
What is the decimal value of this 32-bits two’s
complement number?
1111 1111 1111 1111 1111 1111 1111 1100
(2)
( ) ( ) ( ) ( ) ( ) ( )
1× −231 + 1× 230 + 1× 229 + ... + 1× 22 + 0 × 21 + 0 × 20
= −231 + 230 + 229 + ... + +22 + 0 + 0
= −2147483648 + 2147483644
= −4
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Two's Complement Operations
Negating a two's complement number: invert all bits and add 1
Converting n bit numbers into numbers with more than n bits:
MIPS 16 bit immediate gets converted to 32 bits for arithmetic
copy the most significant bit (the sign bit) into the other bits
0010 -> 0000 0010
1010 -> 1111 1010
"sign extension"
lbu: load byte unsigned
lb: load byte
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•MIPS offers two versions of the set on less than comparison
•有號數的若小於則設定(slt, slti): set on less than, set on less
than immediate
•無號數的若小於則設定(sltu, sltiu): set on less than unsigned,
set on less than immediate unsigned
• 範例．
假設暫存器 $s0 含有以下的二進位數
1111 1111 1111 1111 1111 1111 1111 1111
而暫存器 $s1含有以下的二進位數
0000 0000 0000 0000 0000 0000 0000 0001
經由以下兩個指令運算後，暫存器 $t0及 $t1 的值為何？
slt $t0, $s0, $s1
sltu $t1, $s0, $s1
» 解答
$s0: signed=-1, unsigned=4294967295
$s1: signed=1, unsigned=1
得到 $t0 = 1 及 $t1 = 0.
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EExxaammppllee :: NNeeggaattiioonn SShhoorrttccuutt
Negate 2 , and then check the result by negating -2 .
(10) (10)
2 = 0000 0000 0000 0000 0000 0000 0000 0010
(10)
Negating this number by inverting the bits and adding one.
1111 1111 1111 1111 1111 1111 1111 1101
+
1
1111 1111 1111 1111 1111 1111 1111 1110
= -2
0變1，1變0
0000 0000 0000 0000 0000 0000 0000 0001
+ 1
0000 0000 0000 0000 0000 0000 0000 0010
= 2
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