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1 .How does an assembler work?
In a two-pass assembler
PASS 1: Symbol table generation
PASS 2: Code generation
Illustration of the two passes (follow the class lecture)
.data
L1: .word 0x2345 # some arbitrary value
L2: .word 0x3366 # some arbitrary value
Res: .space 4
.text
.globl main
main: lw $t0, L1($0) #load the first value
lw $t1, L2($0) # load the second value
and $t2, $t0, $t1 # compute the bit-by-bit AND
or $t3, $t0, $t1 # compute the bit-by-bit OR
sw $t3, Res($0) # store result at location in memory
li $v0, 10 # code for program end
syscall
2 .Other architectures
Not all processors are like MIPS.
Example. Accumulator-based machines
A single register, called the accumulator, stores the
operand before the operation, and stores the result
after the operation.
Load x # into accumulator from memory
Add y # add y from memory to the acc
Store z # store acc to memory as z
Can we have an instruction like
add z, x, y # z:= x + y, (x, y, z in memory) ?
For some machines, YES, but not in MIPS! What are the
advantages and disadvantages of such an instruction?
3 .Load-store machines
MIPS is a load-store architecture. Only load and store
instructions access the memory, all other instructions use
registers as operands. What is the motivation?
Register access is much faster than memory access, so
the program will run faster.
Reduced Instruction Set Computers (RISC)
• The instruction set has only a small number of
frequently used instructions. This lowers processor
cost, without much impact on performance.
• All instructions have the same length.
• Load-store architecture.
Non-RISC machines are called CISC (Complex Instruction
Set Computer). Example: Pentium
4 .Another classification
3-address add r1, r2, r3 (r1 ← r2 + r3)
2-address add r1, r2 (r1 ← r1 + r2)
1-address add r1 (to the accumulator)
0-address or stack machines (see below)
Example of stack architecture
Consider evaluating z = x * (y + z)
Push x x x x x * (y+z)
Push y y y+z
Push z z
Add
Multiply
Pop z
5 .Computer Arithmetic
How to represent negative integers? The most widely
used convention is 2’s complement representation.
+14 = 0, 1 1 1 0
-14 = 1, 0 0 1 0
Largest integer represented using n-bits is + (2n-1 – 1)
Smallest integer represented using n-bits is - 2n-1
So, using 4-bits (that includes 1 sign bit),
the largest integer is 0,111 (=7), and
the smallest integer is 1,000 (= -8)
Review binary-to decimal and binary-to-hex conversions.
Review BCD (Binary Coded Decimal) and ASCII codes.
How to represent fractions?
6 .Overflow
+12 = 0, 1 1 0 0 +12 = 0, 1 1 0 0
+2 = 0, 0 0 1 0 +7 = 0, 0 1 1 1
add add
+14 = 0, 1 1 1 0 ? = 1, 0 0 1 1 (WRONG)
Addition of a positive and a negative number does not
lead to overflow. How to detect overflow? Here is a clue.
0 0 0 ⊕ 0 = 0(OK ) 0 1 0 ⊕1 = 1 (NOT OK)
+12 = 0, 1 1 0 0 +12 = 0, 1 1 0 0
+2 = 0, 0 0 1 0 +7 = 0, 0 1 1 1
add add
+14 = 0, 1 1 1 0 ? = 1, 0 0 1 1 (WRONG)
7 .The following sequence of MIPS instructions can detect
overflow in signed addition of $t1 and $t2:
addu $t0, $t1, $t2 # add unsigned
xor $t3, $t1, $t2 # check if signs differ
slt $t3, $t3, $zero # $t3=1 if signs differ
bne $t3, $zero, no_overflow
xor $t3, $t0, $t1 # sum sign = operand sign?
slt $t3, $t3, $zero # if not, then $t3=1
bne $t3, $zero, overflow
no_overflow:
...
...
overflow:
<Do something to handle overflow>
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