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 done else: sub $s0,$s1,$s2 done:

Instructions for Iteration (while) while (save[i] == k) i = i + 1; Let i be in reg $s3 Let k be in reg $s5 Let $t1 have the address of Save array element Loop: 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, a3 Save temp register on the stack Perform operations And return value Restore values stored on the stack Jump back to return address

Steps in Execution of a Procedure Place parameters in a place where procedures can access them Transfer control to the procedure Acquire the storage resources needed for the procedure Perform the desired task Place the result value in a place where the calling program can access it Return control to the point of origin, since a procedure can be called from several points in a program

Register Mapping – Ahah! R0 (r0) = R1 (at) = R2 (v0) = R3 (v1) = R4 (a0) = R5 (a1) = R6 (a2) = R7 (a3) = R8 (t0) = R9 (t1) = R10 (t2) = R11 (t3) = R12 (t4) = R13 (t5) = R14 (t6) = R15 (t7) = R16 (s0) = R17 (s1) = R18 (s2) = R19 (s3) = R20 (s4) = R21 (s5) = R22 (s6) = R23 (s7) = R24 (t8) = R25 (t9) = R26 (k0) = R27 (k1) = R28 (gp) = R29 (sp) = 7fffeffc R30 (s8) = R31 (ra) =

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 call MIPS 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 LIFO Also 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 call Once 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 stack sw $t1,8($sp) # save the temp registers sw $t0,4($sp) sw $so,o($sp) add $to,$ao,$a1 # t0  g + h add $t1,$a2,$a3 # t1  i + j sub $so,$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 stack lw $t0,4(sp) lw $t1,8(sp) addi $sp,$sp,12 jr $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 address It may be used to store arguments It 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 before during after sp fp

Delayed Branches Inst FetchDcd & Op Fetch Execute Branch: Inst Fetch Dcd & Op Fetch Inst Fetch Execute execute successor even if branch taken! Then branch target or 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 bzr4, LL NOP... LL: add rd,...

Delayed Branches li r3, #7 sub r4, r4, 1 bzr4, LL nop LL:sltr1, r3, r5 subir6, r6, 2 li r3, #7 sub r4, r4, 1 bzr4, LL subir6, r6, 2 LL:sltr1, 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:sltr1, r3, r5 li r3, #7 sub r4, r4, 1 bzr4, LL subi r6, r6, 2 Time execute Branch Delay Slot Branch Target ifetchexecute ifetchexecute ifetchexecute

Putting it all together Sort procedure void 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); } }