Representing Numbers
B261 Systems Architecture
Previously
• Computer Architecture– Common misconceptions– Performance
• Instruction Set (MIPS)– How data is moved inside a computer– How instructions are executed
Outline
• Numbers– Limitations of computer arithmetic
• How numbers are represented internally– Representing positive integers – Representing negative integers
• Addition• Bit-wise operations
Binary Representation
• Computers internally represent numbers in binary form
– The sequence of binary digits
»dn-1 dn-2 ... d1 d0 two
– At its simplest, this represents the decimal number which is the sum of terms 2i for each di = 1.
»Eg, 10111two = 16ten+4ten+2ten+1ten = 23ten
» d0 is called the ‘least significant bit’ LSB
» dn-1 is called the ‘most significant bit’ MSB
– But we will need to use different meanings for the bits in order to represent negative or floating-point numbers.
Range of Positive Numbers and Word Lengths
• For an n-bit word length• 2n different numbers can be represented
* including zero.• 0,1,2,..., ((2n) -1) (if only want +ve
numbers)
• A 3-bit word length would support• 0,1,2,3,4,5,6,7
• MIPS has a 32-bit word length• 232 different numbers can be represented• 0,1,2,..., ((232)-1) (= 4,294,967,295).
Negative Numbers
• The range afforded by the word length must simultaneously support both positive and negative numbers, equally.
• Two requirements:– There must be a way, within one word, to tell if it
represents a positive or negative number, and its value.
– A positive number x, and its complement (-x) must clearly add to zero»x + (-x) = 0.
Two’s Complement
• Example of n=3 bit word:• 000 = 0• 001 = 1• 010 = 2• 011 = 3• 100 = -4• 101 = -3• 110 = -2• 111 = -1
For negative numbers the MSB (‘sign bit’) is always 1
For positive numbers the MSB (‘sign bit’) is always 0
Two’s complement
• In general for a n-bit word– There will be positive numbers (and zero)
»0,1,2, … , ((2n-1) - 1)
– Negative numbers»-(2n-1), … , -1
– The highest positive number is ((2n-1) - 1)– The lowest negative numbers is -(2n-1 )– Note the slight imbalance of positive and
negative.
Conversion
• For an n-bit word, suppose that dn-1
is the sign bit.– Then the decimal value is:
» (-dn-1)* (2n-1) + the usual conversion of the remainder of the word.
– Eg, n=3, then»101 = (-1 * 4) + 0 + 1 = -4 + 1 = -3»011 = (0 * 4) + 2 + 1 = 0 + 3 = 3
Negating a Number
• There is a simple two-step process for negation (eg, turn 3 = 011 into -3 = 101):• invert every bit (e.g. replace 011 by 100)• add 1 (e.g. 100+1 = 101).
• Example, 4-bit word 1101• 1101 = -8 + 4 + 0 + 1 = -3
»3 = 0011» Invert bits: 1100»add 1: 1101 = -3 as required.
Sign Extension
• Turning an n-bit representation into one with more bits– Eg, in 3 bits the number -3 is 101– In 4 bits the number -3 is 1101
• Replicate the sign bit from the smaller word into all extra slots of larger word– For 5-bit word from 3 bits: 11101 = -3– For 5-bit word from 4 bits: 11101 = -3
Integer Types
• Some languages (e.g. C++) support two types of int: signed and unsigned.
• Signed integers have their MSB treated as a sign bit– so negative numbers can be represented
• Unsigned integers have their MSB treated without such an interpretation (no negative numbers).
Signed/Unsigned Int
• Suppose we had a 4-bit word, then• int x;
»x could have range 0 to 7 positive»and -8 to -1 negative
• unsigned int x;»x has range 0 to 15 only.
Addition
• Addition is carried out in the same way as decimal arithmetic:
»0101»0001 +»0110
• Subtractions involve negating the relevant number:
»0101 - 0011 =»0101 + 1101 = 0010
Overflow
• Since there are only a fixed set of bits available for arithmetic things can go wrong:
• Consider n=4, and 0110 + 0101 (6+5).
»0110 »0101 +»1011
• But 1011 = -8 + 2 + 1 = -5 !
Overflow Examples
• Consider (6 - (-5))• 0110 - 1011
»= 0110 + 0101»= 1011»= -8 + 3»= -5
• -6 - 3»= 1010 + 1101»= 0111»= 7
Overflow Situations
• A + B– where A>0, B>0, A+B too big, result negative
• A + B– where A<0, B<0, A+B too small, result positive
• A - B– where A>0, B<0, A-B too big, result negative
• A - B– where A<0, B>0, A-B too small, result positive.
Bit Operations
• Shifts– shift a word x bits to the right or left:
»0110 shift left by 1 is 1100»0110 shift right by 1 is 0011
– In C++ shifts are expressed using <<,>>»unsigned int y = x<<5, z = x>>3;»means y is the value of x shifted 5 to left, and
z is x shifted 3 to the right.
– One application is multiplication/division by powers of 2.
AND/OR
• AND is a bitwise operation on two words, where for each corresponding bit a and b:– a AND b = 1 only when a=1 and b=1, otherwise
0.
• OR is a bitwise operation, such that– a OR b = 0 only if both a=0 and b=0, otherwise
1.
• For example:– 1011 AND 0010 = 0010– 1011 OR 0010 = 1011
New MIPS Operations
• add unsigned– addu $1,$2,$3
»operands treated as unsigned ints.
• subtract unsigned– subu $1,$2,$3
»operands treated as unsigned ints
• add immediate unsigned– addiu $1,$2,10
»operands treated as unsigned ints
Exceptions
• Where the signed versions (add, addi, etc) are used– if there is overflow
»then an ‘exception’ is raised by the hardware
»control passes to a special procedure to ‘handle’ the exception, and then returns to the next instruction after the one that raised the exception.
More MIPS
• Load upper immediate– lui $1,100
»$1 = 216 * 100 »(100, shifted left by 16 = upper half of the
word).
• and, or, and immediate (andi), or immediate (ori), shift left logical (sll), shift right logical (srl)– all of form $1,$2,$3 or $1,$2,10 (for
immediate).
Summary
• Basic number representation (integers)
• Representation of negatives• Basic arithmetic (+ and -)• Logical operations• Next time - building an ALU.
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