chapter 2 isa instructions (logical + procedure call)
Post on 28-Dec-2015
233 Views
Preview:
TRANSCRIPT
CHAPTER 2
ISA Instructions (logical + procedure call)
Logical operations
Shift left Example: sll $t2,$so,4 Reg t2 = $so << 4
Shift right Example: srl $t2,$so,4 Reg t2 = $so >> 4
Bit-wise AND Example: and $t0,$t1, $t2 Reg t0 = reg t1 & reg t2
Bit-wise OR Example: or $t0,$t1,$t2 Reg $t0 = $t1 | $t2
Instructions for Selection (if..else)
If (i == j) then f = g + h; else f = g – h; bne $s3,$s4, else add $s0,$s1,$s2 j doneelse: sub $s0,$s1,$s2done:
Instructions for Iteration (while)
while (save[i] == k) i = i + 1;Let i be in reg $s3Let k be in reg $s5Let $t1 have the address of Save array elementLoop: sll $t1,$s3,2 add $t1,$t1$s6 lw $t0,0($t1) bne $t0,$s5,Exit addi $s3,$s3,1 j Loop
Compiling C procedures
int leaf_example (int g, int h, int i, int j){ int f; f = (g + h) – (i + j); return f;}
How do you pass the parameters? How does compiler transport the parameters?
Passing Parameters/arguments
Special registers for arguments: a0, a1, a2, a3Save temp register on the stackPerform operations And return valueRestore values stored on the stackJump back to return address
Steps in Execution of a Procedure
Place parameters in a place where procedures can access them
Transfer control to the procedureAcquire the storage resources needed for the
procedurePerform the desired taskPlace the result value in a place where the
calling program can access itReturn control to the point of origin, since a
procedure can be called from several points in a program
Register Mapping – Ahah!
R0 (r0) = 00000000R1 (at) = 00000000R2 (v0) = 00000000R3 (v1) = 00000000 R4 (a0) = 00000000R5 (a1) = 00000000R6 (a2) = 00000000R7 (a3) = 00000000R8 (t0) = 00000000R9 (t1) = 00000000R10 (t2) = 00000000R11 (t3) = 00000000R12 (t4) = 00000000R13 (t5) = 00000000R14 (t6) = 00000000R15 (t7) = 00000000
R16 (s0) = 00000000R17 (s1) = 00000000 R18 (s2) = 00000000 R19 (s3) = 00000000 R20 (s4) = 00000000R21 (s5) = 00000000 R22 (s6) = 00000000 R23 (s7) = 00000000 R24 (t8) = 00000000R25 (t9) = 00000000R26 (k0) = 00000000R27 (k1) = 00000000 R28 (gp) = 10008000R29 (sp) = 7fffeffcR30 (s8) = 00000000R31 (ra) = 00000000
Procedure Call Conventions
$a0-$a3 : four argument registers in which to pass parameters
$vo-$v1: two value registers in which to return values$ra: one return address register to return to point of
origin of the callMIPS assembly language also has a special instruction
jal (jump and link) that saves the return address in $ra before transferring control to the procedure. Eg: jal ProcedureAddress
Another instruction “jr” transfers control back to the called location. Eg. jr $ra
Using the Stack “automatic storage”
Automatic store for “workspace” and temporary registers.
Use the stack: Use the stack pointer to access stack; Remember stack operates on LIFOAlso note that $ZERO is a convenience
register that stores the value 0 (check your green sheet attached to your textbook)
Now we are ready to translate the procedure:
Translating the procedure
int leaf_example (int g, int h, int i, int j){ int f; f = (g + h) – (i + j); return f;}Parameters g, h, i, j will be stored in argument
registers: $a0,$a1,$a2,$a3 before the callOnce inside we plan to use $s0, $t0, $t1; so we
need save their values on the stack;Then use them to compute (g+h), (i+j) and f
MIPS code
addi $sp,$sp,-12 # make room on the stacksw $t1,8($sp) # save the temp registerssw $t0,4($sp)sw $so,o($sp)
add $to,$ao,$a1 # t0 g + hadd $t1,$a2,$a3 # t1 i + jsub $so,$t0,$t1 # f = t0 – t1
add $v0,$so,$zero # returns f ($vo = $s0 + 0)
lw $s0,0(sp) # restore saved values into temp registers from stacklw $t0,4(sp)lw $t1,8(sp)addi $sp,$sp,12jr $ra # jump back to the return address
Allocating Space on Stack
Stack is used to save and restore registers when calling a procedure
It may also be used to store return addressIt may be used to store argumentsIt may be used to store local arrays and data The segment of the stack containing a
procedure’s saved registers and local variables is called “procedure frame” or “activation record”
A frame pointer ($fp) points to first word of the frame of a procedure.
Stack and frame pointers
sp
fp
Saved arguments
Saved return addr
Saved regs.Locals
arrays& structures
fp
sp
beforeduring after
spfp
Delayed Branches
Inst Fetch Dcd & Op Fetch ExecuteBranch:
Inst Fetch Dcd & Op Fetch
Inst Fetch
Executeexecute successoreven if branch taken!
Then branch targetor continue Single delay slot
impacts the critical path
•Compiler can fill a single delay slot with a useful instruction 50% of the time.
• try to move down from above jump
•move up from target, if safe
add r3, r1, r2
sub r4, r4, 1
bz r4, LL
NOP
...
LL: add rd, ...
Delayed Branches
li r3, #7
sub r4, r4, 1
bz r4, LL
nop
LL: slt r1, r3, r5
subi r6, r6, 2
li r3, #7
sub r4, r4, 1
bz r4, LL
subi r6, r6, 2
LL: slt r1, r3, r5
compiler
Branch and Pipelines
By the end of Branch instruction, the CPU knows whether or not the branch will take place.
However, it will have fetched the next instruction by then, regardless of whether or not a branch will be taken.
Why not execute it?
ifetch
LL: slt r1, r3, r5
li r3, #7
sub r4, r4, 1
bz r4, LL
subi r6, r6, 2
Time
execute
Branch
Delay Slot
Branch Target
ifetch execute
ifetch execute
execute
ifetch execute
Putting it all together
Sort procedurevoid sort (int v[], int n){ int i, j; for (i = 0; i < n; i = i+1) { // outer loop for (j = i - 1; j >= 0 && v[j] > v[j+1]; j = j-1)
{ swap(v, j); }} }
top related