lecture 4: advanced instructions, control, and branching cont. een 312: processors: hardware,...

Lecture 4: Advanced Instructions, Control, and Branching cont.

EEN 312: Processors: Hardware, Software, and Interfacing

Department of Electrical and Computer Engineering

Spring 2014, Dr. Rozier (UM)

COURSE PLAN FOR TODAY

1.Control Instructions2.Basic Memory Access

CONTROL INSTRUCTIONS

Branching

• Branches allow us to transfer control of the program to a new address.– b (<suffix>) <label>– bl (<suffix>) <label>

b startbl start

Branch (b)

• Branch, possibly conditionally, to a new address.

beq subroutine @ If Z=1, branch

• Good practice to use bal instead of b.

Branch with link (bl)

• Branch, possibly conditionally, to a new address.– Before the branch is complete, store the PC in the

LR.– Allows easy return from the branch.

bleq subroutine @ If Z=1, branch, saving the PC

Branch with link (bl)

• How do we get back once we’ve saved the PC?

mov pc, lr

• Moves the contents of the link register to the program counter.

Goto vs. Subroutine

• b – Allows us to transfer control of the program without returning.

• bl – Allows us to transfer control of the program, with returning.

Implementing If Statements

• C code:if (i == j) f = g+h;else f = g - h;

• ARM code cmp r0, r1 @ Set flags via r0-r1 and discard

bne Elseadd r2, r3, r4 @ r2 = r3 + r4bal Exit

Else:sub r2, r3, r4 @ r2 = r3 - r4

Exit:

Implementing Loop Statements

• C code:while (i < j) i += 1;

• ARM code Loop:

cmp r0, r1bge Exitadd r0, r0, #1bal Loop

Exit:

i < j?i < j?

i=i+1i=i+1

i<j

ExitExit

i>=j

What about a Case Statement?

• Say we have a case statement:switch(x) {

case 0: foo(); break;case 1: bar(); break;case 2: baz(); break;case 3: qux(); break;

}

Jump Tables

• Set up a portion of memory as such:

• If our case variable is stored in r0…– r0 << #2 is the index into our jump table of the

function address.

Memory Location Contents

0x???? + 0 address of foo

0x???? + 4 address of bar

0x???? + 8 address of baz

0x???? + 12 address of qux

Cases Statements as Jump Tables

(assume r0 holds the switch variable)

ldr r1, =jumptableldr pc, [r1, r0, lsl #2]

Wasn’t that easy?

Jump Table Exercise

• Convert the follow C-style code into ARMv6 instructions:

x = getPrime();switch(x) {

case 1: firstPrime(); break;case 3: secondPrime(); break;case 5: thirdPrime(); break;case 7: fourthPrime(); break;default: notFirstFourPrimes();

}

Pseudo-Instructions

• Notice the use of:– ldr r0, =casetable– We saw this before in helloworld.s

• What is really going on here?

Pseudo-Instructions

Code as we wrote it:

Disasembled code:

ldr r1, =string

swi 0

mov r7, #1

swi 0

0x8080 ldr r1, [pc, #8]

0x8084 svc 0x0

0x8088 mov r7 #1

0x808c svc 0x0

0x8090 muleq r1 r4 r0

This is weird…

• Let’s play with gdb…x/x 0x80900x8090 <_exit+8>: 0x00010094x/x 0x100940x10094 <string>: “Hello World!\nA\025”

Looking back at the assembled code…

• While at instruction 0x8080p/x $pc0x8080

0x8080 ldr r1, [pc, #8]

0x8084 svc 0x0

0x8088 mov r7 #1

0x808c svc 0x0



• While at instruction 0x8080p/x $pc+80x8088

0x8080 ldr r1, [pc, #8]

0x8084 svc 0x0

0x8088 mov r7 #1

0x808c svc 0x0


The real value of $pc

+4 +4


• While at instruction 0x8080p/x $pc+80x8088p/x ($pc+8)+80x8090

0x8080 ldr r1, [pc, #8]

0x8084 svc 0x0

0x8088 mov r7 #1

0x808c svc 0x0


So why does it show up as muleq?

• Representing instructions

– Condition Field• 0000 – EQ

– 0000 | 000000 | 0 | 0 |????|????|????| 1001|????

mul r1, r4, r0mul{<cond>}{S} rd, rm, rs

Cond 000000 A S Rd Rn Rs 1001 Rm

Instruc 0000 000000 0 0 ???? ???? ???? 1001 ????

Hex 0 0 0 1 0 0 9 4

Bin 0000 0000 0000 0001 0000 0000 1001 0100

So why does it show up as muleq?

• Representing instructions

mul r1, r4, r0mul{<cond>}{S} rd, rm, rsmul 0001, 0100, 0000

Cond 000000 A S Rd Rn Rs 1001 Rm

Instruc 0000 000000 0 0 ???? ???? ???? 1001 ????

Hex 0 0 0 1 0 0 9 4

Bin 0000 0000 0000 0001 0000 0000 1001 0100

So what is this?

Code as we wrote it:

Disasembled code:

ldr r1, =string

swi 0

mov r7, #1

swi 0

0x8080 ldr r1, [pc, #8]

0x8084 svc 0x0

0x8088 mov r7 #1

0x808c svc 0x0


The problem with immediates

• The fact that instructions, AND all their arguments, must take up only 32 bits limits the size of immediates to 1 byte.– Range 0 – 255.– Hello world was in 0x10094– PC was at 0x8088– Max offset with immediate value?

• 0x8088 + 0xFF = 0x8187

Enter, the Literal Pool0x8080 ldr r1, [pc, #8]

0x8084 svc 0x0

0x8088 mov r7 #1

0x808c svc 0x0

0x8090 00 01 00 94

Last instruction in basic block

Literal Pool

Basic Blocks• A basic block is a sequence of instructions with

– No embedded branches (except at end)– No branch targets (except at beginning)

A compiler identifies basic blocks for optimization

An advanced processor can accelerate execution of basic blocks

So how do we use the Literal Pool?

• We’ve been using it all along.

ldr r0, =string

ldr r0, [pc, #8]

What about procedures?

• Implement the following:

foo(int x) {x = x + 1

}main() {x = 1;foo(x);}

What about procedures?

• Implement the following:

int foo(int x) {if (x < 5)

x = foo(x+1);return(x)

}main() {x = 1;x = foo(x);}

The problem with Procedures

• When we never branch back, who cares what state the registers are in?

• If we branch and return, we expect our registers to be in the state that we left them!– Procedures need to use registers too!

A Call Chain

Image by David Thomas

Procedure Calling

1. Place parameters for procedure in registers2. Transfer control to procedure3. Procedure acquires storage4. Procedure performs function.5. Procedure places return value in appropriate

register6. Return control

The Stack

• Region of memory managed with stack discipline.

• Accessed with pushand pop

The Stack

Procedure Call Example0x0000x0040x008

push {lr}bl procedure_2pop {pc}

0x100

0x104

0x108

0x10c

0x110

0x114

0x118

lr 0x080

sp 0x100

pc x



0x100 0x080

0x104

0x108

0x10c

0x110

0x114

0x118

lr 0x080

sp 0x104

pc x



0x100 0x080

0x104 0x008

0x108 0xa4f

0x10c 0xfff

0x110 0xcc4

0x114 0xbeef

0x118 0xdead

lr 0x008

sp 0x104

pc x



0x100 0x080

0x104 0x008

0x108 0xa4f

0x10c 0xfff

0x110 0xcc4

0x114 0xbeef

0x118 0xdead

lr 0x080

sp 0x100

pc x

Stack Frames

• Stack frames “belong” to a procedure.

• Store local variables here (they go out of scope automatically)

• Can communicate with other procedures with a stack frame.

Communicating with Procedures

0x100 lr

0x104 return 0

0x108 return 1

0x10c

0x110

0x114

0x118

Caller sets up stack frame

Communicating with Procedures

0x100 lr

0x104 return 0

0x108 return 1

0x10c

0x110

0x114

0x118

Callee stores values before return

Procedures and Register Use

• What if we want to use some registers in the procedure?

• Caller could have data in it!

Stack Discipline and ABI

• Stack Discipline is important

• Define an Application Binary Interface– How should procedures communicate? How

should the be called?

• Consistency, following standards, is the key.

Conventions and Discipline

Caller• Caller saves temporary

values in its frame before the call.

• Caller restores values in its frame after the call.

Callee• Callee saves temporary

values in its frame before using.

• Callee restores values in its frame after using.

WRAP UP

For next time

•More on stacks, procedures, and calling.

lecture 4: advanced instructions, control, and branching cont. een 312: processors: hardware,...

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