10/4: lecture topics

31
10/4: Lecture Topics • Overflow and underflow • Logical operations • Procedure calls

Upload: rian

Post on 08-Jan-2016

15 views

Category:

Documents


1 download

DESCRIPTION

10/4: Lecture Topics. Overflow and underflow Logical operations Procedure calls. Overflow and Underflow. Overflow occurs when a number is too big to represent usually as the result of a numerical operation unsigned ints, > 2 32 -1 signed ints, > 2 31 -1 floats, > 3.40282347e+38 - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: 10/4: Lecture Topics

10/4: Lecture Topics

• Overflow and underflow• Logical operations• Procedure calls

Page 2: 10/4: Lecture Topics

Overflow and Underflow• Overflow occurs when a number is too big to

represent usually as the result of a numerical operation– unsigned ints, > 232-1– signed ints, > 231-1– floats, > 3.40282347e+38– doubles, > 1.7976931348623157e+308

• Underflow means the number is too small to represent– unsigned ints < 0– signed ints, < -231

– floats, > 1.17549435e-38– doubles, > 2.2250738585072014e-308

Page 3: 10/4: Lecture Topics

Logical Operations• Bitwise operations (and/andi & or/ori)

– and 0110 0011 & 1100 0110 = 0100 0010– or 0110 0011 | 1100 0110 = 1110 0111

• Shift operations – left shift 0000 1111 << 2 = 0011 1100 (sll)– right shift 0000 1111 >> 2 = 0000 0011

• Signed >>– 1110 0010 >> 2 = 1111 1000– sra

• Unsigned >> – 1110 0010 >> 2 = 0011 1000– srl

Page 4: 10/4: Lecture Topics

Examples

• Evaluate the following– (1100 1100 & 0001 1111) | 1100 0000– (1100 1100 >> 3) | 1100 0000

• Fill in the body of this procedure

int GetBitFromPosition( int num, int pos ) {if( ( pos < 0 ) || ( pos >= 32 ) ) {

fprintf( stderr, “You idiot.\n” );return 0;

}

return }

Page 5: 10/4: Lecture Topics

Examples Continued

• f is a single precision floating point number write code to extract the actual exponent from f

Page 6: 10/4: Lecture Topics

Procedure calls in assembly

int fact( int n ) { int result;

if( n <= 1 ) result = 1; else result = n * fact(n-1);

return result;}

main() {int i;i = fact( 5 );

}

Page 7: 10/4: Lecture Topics

Procedure Call

• More than just a branch and a return

• Data goes in– arguments, parameters

• Data goes back out– return value

• What makes this possible?– the stack

Page 8: 10/4: Lecture Topics

Review of Stacks

• Two operations:– push an item

onto the stack– pop an item off

the stack

Page 9: 10/4: Lecture Topics

Stack Implementations

• Pretty easy to do with a linked list– You probably saw this in 143 or 373

Top

Page 10: 10/4: Lecture Topics

A Stack in an Array

• Linked lists are nice if you have them

• Arrays are a lot faster

A[0]

A[1]

A[2]

A[3]

A[4]

A[5]

Top

Page 11: 10/4: Lecture Topics

Calling Conventions

• Sequence of steps to follow when calling a procedure

• Determines:– where arguments are passed to the

callee– how to transfer control from caller to

callee and back– where return values passed back out– no unexpected side effects

• such as overwritten registers

Page 12: 10/4: Lecture Topics

Calling Conventions

• Mostly governed by the compiler• We’ll see a MIPS calling convention

– Not the only way to do it, even on MIPS

– Most important: be consistent

• Procedure call is one of the most unpleasant things about writing assembly for RISC architectures

Page 13: 10/4: Lecture Topics

A MIPS Calling Convention

1. Place parameters where the procedure can get them

2. Transfer control to the procedure3. Get the storage needed for the

procedure4. Do the work5. Place the return value where the

calling code can get it6. Return control to the point of origin

Page 14: 10/4: Lecture Topics

Step 1: Parameter Passing

• The first four parameters are easy - use registers $a0, $a1, $a2, and $a3

• You’ve seen this already• What if there are more than four

parameters?

Page 15: 10/4: Lecture Topics

Step 2: Transfer Control

• Getting from caller to callee is easy -- just jump to the address of the procedure

• Need to leave a way to get back again

• Special register: $ra (for return address)

• Special instruction: jal

Page 16: 10/4: Lecture Topics

Jump and Link

Calling code

Procedure

jal proc

proc: add ..

Page 17: 10/4: Lecture Topics

Step 3: Acquire Storage

• What storage do we need?– Registers– Other local variables

• Where do we get the storage?– From the stack

Page 18: 10/4: Lecture Topics

Refining Program Layout

Address0

0x00400000

0x10000000

0x10008000

0x7fffffff

Reserved

Text

Static data

Stack

Program instructions

Global variables

Dynamic data heap

Local variables,

saved registers

Page 19: 10/4: Lecture Topics

Saving Registers on the Stack

$sp

$sp

$sp

Before Procedure

$s0$s1$s2

During Procedure

After Procedure

Page 20: 10/4: Lecture Topics

Assembly for Saving Registers

• We want to save $s0, $s1, and $s2 on the stack

sub $sp, $sp, 12 # make room for 3 words # “addi $sp, $sp, -12”sw $s0, # store $s0sw $s1, # store $s1sw $s2, # store $s2

Page 21: 10/4: Lecture Topics

Step 4: Do the work

• We called the procedure so that it could do some work for us

• Now is the time for it to do that work

• Resources available:– Registers freed up by Step 3– All temporary registers ($t0-$t9)

Page 22: 10/4: Lecture Topics

Callee-saved vs. Caller-saved

• Some registers are the responsibility of the callee– callee-saved registers– $s0-$s7

• Other registers are the responsibility of the caller– caller-saved registers– $t0-$t9

Page 23: 10/4: Lecture Topics

Step 5: Return values

• MIPS allows for two return values• Place the results in $v0 and $v1• You’ve seen this too• Why are there two return values?• What if the procedure needs more

than two return values?

Page 24: 10/4: Lecture Topics

Step 6: Return control

• Because we laid the groundwork in step 2, this is easy

• Address of the point of origin + 4 is in register $ra

• Just use jr $ra to return

Page 25: 10/4: Lecture Topics

An Example

int leaf(int g, int h, int i, int j) { int f;

f = (g + h) - (i + j); return f;}

Let g, h, i, j be passed in $a0, $a1, $a2, $a3, respectively

Let the local variable f be stored in $s0

Page 26: 10/4: Lecture Topics

Compiling the Exampleleaf: sub $sp, $sp, 4 # make room for $s0 # addi $sp, $sp, -4 sw $s0, 0($sp) # store $s0 add $t0, $a0, $a1 # $t0 = g + h add $t1, $a2, $a3 # $t1 = i + j sub $s0, $t0, $t1 # $s0 = f add $v0, $s0, $zero # copy result lw $s0, 0($sp) # restore $s0 addi $sp, $sp, 4 # put $sp back jr $ra # jump back to caller

Page 27: 10/4: Lecture Topics

Nested Procedures

• Suppose we have code like this:

• Potential problem: the return address is stored in $ra which will get overwritten

main() { foo();}

int foo() { return bar();}

int bar() { return 6;}

Page 28: 10/4: Lecture Topics

A Trail of Bread Crumbs

• The registers $s0-$s7 are not the only ones we save on the stack

• What can the caller expect to have preserved across procedure calls?

• What can the caller expect to have overwritten during procedure calls?

Page 29: 10/4: Lecture Topics

Preservation Conventions

Preserved Not PreservedSaved registers:

$s0-$s7

Stack pointer register: $sp

Return address register: $ra

Stack above the stack pointer

Temporary registers: $t0-$t9

Argument registers: $a0-$a3

Return value registers: $v0-$v1

Stack below the stack pointer

Page 30: 10/4: Lecture Topics

A Brainteaser in C

• What does this program print? Why?

#include <stdio.h>

int* foo() { int b = 6; return &b;}

void bar() { int c = 7;}

main() { int *a = foo(); bar(); printf(“The value at a is %d\n”, *a);}

Page 31: 10/4: Lecture Topics

Activation Record

• For a procedure call, the activation record is the portion of the stack containing – saved registers– local variables

• Also known as procedure frame