Post on 02-Nov-2019
CPU CPU Mem Mem
CISC RISC
add eax, [ebx] add[ecx],eax
ld r1, [r2] add r1,r1,r3 add r1,r1,r4 st r1 [r5]
add eax, [1] add[2],eax
ld r1, [1] add r1,r1,r2 add r1,r1,r3 st r1 [2]
Example
add r3,r1,r1 add rd,rs,rt
str r6,[r1 + 1] str rd, [rs + c]
st r0, [r1] st rd, [rs]
ldr r2,[r3 + 4] ldr rd, [rs + c]
ld r2,[4] ld rd, [rs]
ldi r0,3 ldi rd,C
Let’s build a Computer
Let’s take a RISC. What do we need ?
• Memory
• Registers
• ALU
• Control Circuits
• A programming language
• A good Name - Simple Although Meaningful
Here’s Sam-RISC
Data
Memory
Instruction reg
Code
Memory ALU
r1
r2
r0
X
Y
W
X Y
W
0
1
7 mar
mdr
The Instruction Register
001010 00010 00001 00011 unused Code
Memory
Add r2,r1,r3
add 2 1 3
3 1 2
Loaded with the instruction, the IR decodes this into bits which drive the
CPU digital logic circuits
add
Electronic Wires
Control Path
001010 00010 00001 00011 unused
000101 00010 00001 00011 unused
add r2, r1, r3
sub r2, r1, r3
ALU
ALU
+
+
-
-
The add instruction is decoded and produces digital signals which select the + function in the ALU Add !
Subtract !
The sub function decoded produces different digital signals
r1 r3
r1 r3
r2
r2
Sam-RISC and ARM-Cortex are 32 bit
001010 00110 01001 00011 unused
001010 00110
001010 00010 00001 0101001111111011
opcode rd rs rt unused
opcode rd rs c (16-bit address)
add rd,rs,rt
ldr rd,[rs+c]
ldr rd,[c]
opcode rd c (21-bit address)
32 bits wide
‘Load Immediate’
Constant C rd ldi
opcode destination
In load immediate we get the constant C immediately following the opcode into the reg.
ldi r9 , 5
5
5 r9
All reference to memory has
gone!
Load ‘5’ straight into r9
unused
A simple ‘Load’ instruction
‘Load into rd the contents of memory at address which is in reg rs.’ Simple!
7
69 6
231 5
115 4
145 3
2
1
0
ldr r9 , [r1]
3
145 r9
145
rs rd ld
opcode destination
Single source reg
1. Let’s say have already
loaded r1 with 3 2. Get data from mem at addr r1 (=3)
2. Load the data into r9
memory
A more complex ‘Load’
constant c rs rd ldr
opcode destination
Source
Load register rd with the contents of memory which you find at address r1 + c.
7
69 6
231 5
115 4
145 3
2
1
0
ldr r9 , [r1 + 2]
3 + 2
5
231 r9
231
The mem address
is formed as a sum
memory