CHAPTER 3 Arithmetic For Computers 1/31/2009 1

Topics for discussion 1/31/ Number system: { radix/base, a set of distinct digits, operations} Radix conversion ASCII versus binary representation Signed arithmetic; sign extension Bounds check; validation Addition and subtraction algorithms Multiplication algorithms Floating point representation Floating point arithmetic algorithms

Number system 1/31/ Radix or Base: 10 for decimal system, 2 for binary system, 8 for octal, 12 for duo-decimal (to count dozens), 16 for hexa-decimal Decimal digits: {0,1,2,3,4,5,6,7,8,9} Binary {0,1}: binary digit is a “bit” Octal {0,1,2,3,4,5,6,7} Hex {0,..9, A, B, C, D, E, F} If we assume digits are number from right to left starting from 0 th digit as the least significant digit (Little Endian), the value of the i th digit d in is: d x base i Example: Your pay is $101 per hour. Would you prefer it (the base) in decimal, octal, hexadecimal or binary?

Number Representation 1/31/ On the keyboard it is represented by ASCII: American Standard Code for Information Interchange: 7 bit code represented by a byte container. For character representation this is a nice system. How efficient is this representing numbers for processing? Consider 4 ASCII digits. What is the range of integers you can represent with this?  0 – 9999  4 X 8 = 32 bits With 32 bits and binary systems and only positive numbers:  0 – ( ). What is this value? Approx: 4,000,000,000 or 4G !

Signed Numbers 1/31/ When we allow negative and positive numbers, half the range is occupied by positive numbers and the other by negative numbers. How to represent the sign? Using a bit ? 0 for positive and 1 for negative? Then with 32 bits: + 0 to +( ) positive range - 0 to –( ) negative range A better representation for negative number is 2’s complement. How to compute 2’s complement? What’s the advantage of 2’s complement?  Subtraction is equivalent to adding 2’s complement of the second operand.  Sign extension can be used to extend the number from, from 16 bits to 32 bits for example.

Dealing with Overflow 1/31/ OperationOperand AOperand BResult indicating overflow A + B>= 0 < 0 A + B< 0 > = 0 A - B>= 0< 0 A - B< 0>= 0> 0

Dealing with overflow 1/31/ Signed operations that result in overflow cause an exception whereas unsigned operations on overflow do not cause exception. Example:  add, addi, sub will cause exception on overflow  addu, addiu, and subu will not cause exception on overflow How to detect overflow in unsigned then? addu $t0,$t1,$t2 xor $t3,$t1,$t2 # check if signs differ slt $t3,$t3,$zero # signs differ? bne $t3,$zero,No_Overflow # signs differ nop # signs area same xor $t3,$t0,$t1 # find sign of sum slt $t3,$t3,$zero # if sum sign is diff bne $t3,$zero,Overflow

Branches and Jumps 1/31/ beq $s1,$s2,lab1 # if $s1 == $s2 jump to label lab1 bne $s1,$s2,lab2 # if $s1 ≠ $s2 jump to label lab2 j lab3 # unconditional jump to lab3 jal proc3 # jump and link to proc3 jr retAddr # jump back to callee beqz $s2, lab4 # is $s2 == 0 then jump to lab4