Difference between revisions of "CSC231 Homework 8 Solution 2012"
(→Option 3) |
(→Option 3) |
||
| Line 759: | Line 759: | ||
<blockquote> | <blockquote> | ||
Who controls the default maximum stack size; OS or compiler? | Who controls the default maximum stack size; OS or compiler? | ||
| − | + | <br /> | |
The compiler typically. The OS/hardware does limit it to a certain extent. Default is 8MB on linux IIRC. Think of ulimit -s on Linux (to change stack sizes). | The compiler typically. The OS/hardware does limit it to a certain extent. Default is 8MB on linux IIRC. Think of ulimit -s on Linux (to change stack sizes). | ||
| − | + | <br /> | |
Is the default maximum scaled according to total memory? (ie a machine with 2gb memory would have larger default size than a machine with only 512mb) For this example both machines are same os/compiler setup, just different amounts of system RAM. | Is the default maximum scaled according to total memory? (ie a machine with 2gb memory would have larger default size than a machine with only 512mb) For this example both machines are same os/compiler setup, just different amounts of system RAM. | ||
| − | + | <br /> | |
No. Until and unless you do it yiurself.You can alter stack sizes via compiler switches. | No. Until and unless you do it yiurself.You can alter stack sizes via compiler switches. | ||
| − | + | <br /> | |
| − | + | <code><pre> ld --stack=<STACK_SIZE></pre></code> | |
| − | + | <br /> | |
or | or | ||
| − | + | <br /> | |
| − | + | <code><pre> gcc -Wl,--stack=<STACK_SIZE></pre></code> | |
| − | + | <br /> | |
The C++ Standard's take on the issue of stacks and heaps: | The C++ Standard's take on the issue of stacks and heaps: | ||
| − | + | <br /> | |
The standard is based on an abstract machine and does not really concern itself with hardware or stacks or heaps. It does talk about an allocated store and a free store. The free store is where you'd be if you are calling new (mostly). FWIW, an implementation can have only one memory area masquerading as both stack and heap when it comes to object allocation. | The standard is based on an abstract machine and does not really concern itself with hardware or stacks or heaps. It does talk about an allocated store and a free store. The free store is where you'd be if you are calling new (mostly). FWIW, an implementation can have only one memory area masquerading as both stack and heap when it comes to object allocation. | ||
| − | + | <br /> | |
Your question, therefor, boils down to be an implementation specific issue rather than a language issue. | Your question, therefor, boils down to be an implementation specific issue rather than a language issue. | ||
| + | <br /> | ||
| − | |||
</blockquote> | </blockquote> | ||
Revision as of 08:47, 14 November 2012
--D. Thiebaut 12:15, 13 November 2012 (EST)
Contents
Problem #1, Version 1
This version contains the complete code. Version 2 below separates the dumpRegs functions and main program into two files.
;;; ---------------------------------------------------------
;;; hw8aSol.asm
;;; D. Thiebaut
;;; 11/13/12
;;; This program supports several functions that allow the user
;;; to dump the contents of the main 6 registers (eax, ebx, ecx,
;;; edx, esi and edi) to the screen. Great for debugging without
;;; a debugger.
;;;
;;; Typical output:
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;;
;;; To assemble, link and run:
;;; nasm -f elf hw8aSol.asm
;;; ld -melf_i386 hw8aSol.o -o hw8aSol
;;; ./hw8aSol
;;; ---------------------------------------------------------
section .data
hex db "0123456789ABCDEF"
regs db "axbxcxdxsidi"
line db "+-----------------+", 10
string db "| exx = .... .... |", 10
lineLen equ $-string
reg equ string+3
word1 equ string+8
section .text
global _start
_start:
;;; Initialize the registers
mov eax, 0x12345678
mov ebx, 0x55FF55FF
mov ecx, 0xFEDCBA98
mov edx, 0x00000000
mov esi, 0x11112222
mov edi, 0x22223333
;;; dump them twice to verify that no registers gets modified...
call dumpRegs
call dumpRegs
;;; exit back to OS
mov ebx, 0
mov eax, 1
int 0x80
;;; ---------------------------------------------------------
;;; dumpRegs: dumps the contents of the 6 main registers (eax
;;; ebx, ecx, edx, esi and edi) in a box on the screen.
;;; the screen. Does not modify any register
;;; ---------------------------------------------------------
dumpRegs: pushad
;;; put all the registers to print in the stack. The loop
;;; will pop each one, one at a time, once per loop.
push edi
push esi
push edx
push ecx
push ebx
push eax
;;; print top bar
call printBar
;;; prepare to loop 6 times (6 registers )
mov ecx, 6
mov esi, 0
.for pop ebx ;get value of register to convert from stack
push ecx ;save loop counter
lea eax, [regs + esi*2] ;get address of string representing register
mov ax, word [eax] ;ax gets "ax" or "bx" or "cx" or ... "di"
mov word[reg], ax ;modify string with name of register
mov edx, word1 ;edx points to buffer where to store hex digits
;ebx contains register to convert
call storeDWord ;store hex equivalent of ebx in string
mov ecx, string ;print the whole string
mov edx, lineLen
call printInt80
inc esi ;index into register names incremented
pop ecx ;get loop counter back
loop .for
;;; print bottom bar
call printBar
popad
ret
;;; ---------------------------------------------------------
;;; storeDWord: gets
;;; ebx: register to convert
;;; edx: address where to put information
;;; ---------------------------------------------------------
storeDWord: rol ebx, 16 ;bring most sig word in least sig word
call storeWord
inc edx ;make edx skip space in string
rol ebx, 16 ;get lower byte back in least sig pos
call storeWord
ret
;;; ---------------------------------------------------------
;;; storeWord: gets
;;; bx: word to convert
;;; edx: address where to put information. edx incremented by 4
;;; ---------------------------------------------------------
storeWord: rol bx, 8 ;put most sig byte of bx in lower position
call storeByte ;and convert it. Store it in where edx
;points to.
rol bx, 8 ;same with lower nybble
call storeByte
ret
;;; ---------------------------------------------------------
;;; storeByte: gets
;;; bl: byte to convert
;;; edx: address where to store information. edx incremented by 2 automatically
;;; ---------------------------------------------------------
storeByte: push ebx
push eax
rol bl, 4 ;get upper nybble in lower position (LSB)
mov al, bl ;play with al now
and eax, 0x0000000F ;clear all but the least significant nybble
add eax, hex ;get ascii in hex array equivalentn to nybble
mov al, [eax] ;hex digit now in bl
mov byte [edx], al ;store ascii in string
inc edx
rol bl, 4 ;get lower nybble back
mov al, bl ;play with al
and eax, 0x0000000F
add eax, hex
mov al, [eax]
mov byte [edx], al
inc edx
pop eax
pop ebx
ret
;;; ---------------------------------------------------------
;;; printInt80: prints on the screen the string whose address
;;; is in ecx and length in edx.
;;; does not modify registers
;;; ---------------------------------------------------------
printInt80: push eax
push ebx
mov eax, 4
mov ebx, 1
int 0x80
pop ebx
pop eax
ret
;;; ---------------------------------------------------------
;;; printBar: prints the horizontal bar
;;; does not modify registers
;;; ---------------------------------------------------------
printBar: push ecx
push edx
mov ecx, line
mov edx, lineLen
call printInt80
pop edx
pop ecx
ret
Problem #1, Version 2
Main Program
;;; ---------------------------------------------------------
;;; hw8aSol2.asm
;;; D. Thiebaut
;;; 11/13/12
;;; This program uses the library dumpRegs.asm to dump
;;; the contents of the main 6 registers (eax, ebx, ecx,
;;; edx, esi and edi) to the screen. Great for debugging without
;;; a debugger.
;;;
;;; Typical output:
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;;
;;; To assemble, link and run:
;;; nasm -f elf hw8aSol.asm
;;; ld -melf_i386 hw8aSol.o -o hw8aSol
;;; ./hw8aSol
;;; ---------------------------------------------------------
%include "dumpRegs.asm"
section .text
global _start
_start:
;;; Initialize the registers
mov eax, 0x12345678
mov ebx, 0x55FF55FF
mov ecx, 0xFEDCBA98
mov edx, 0x00000000
mov esi, 0x11112222
mov edi, 0x22223333
;;; dump them twice to verify that no registers gets modified...
call dumpRegs
call dumpRegs
;;; exit back to OS
mov ebx, 0
mov eax, 1
int 0x80
Library
;;; ---------------------------------------------------------
;;; dumpRegs.asm
;;; D. Thiebaut
;;; 11/13/12
;;; This libary supports several functions that allow the user
;;; to dump the contents of the main 6 registers (eax, ebx, ecx,
;;; edx, esi and edi) to the screen. Great for debugging without
;;; a debugger.
;;;
;;; Typical output:
;;; +-----------------+
;;; | eax = 1234 5678 |
;;; | ebx = 55FF 55FF |
;;; | ecx = FEDC BA98 |
;;; | edx = 0000 0000 |
;;; | esi = 1111 2222 |
;;; | edi = 2222 3333 |
;;; +-----------------+
;;; Include in program as follows:
;;;
;;;
;;; %include "dumpRegs.asm"
;;; ---------------------------------------------------------
section .data
hex db "0123456789ABCDEF"
regs db "axbxcxdxsidi"
line db "+-----------------+", 10
string db "| exx = .... .... |", 10
lineLen equ $-string
reg equ string+3
word1 equ string+8
section .text
;;; ---------------------------------------------------------
;;; dumpRegs: dumps the contents of the 6 main registers (eax
;;; ebx, ecx, edx, esi and edi) in a box on the screen.
;;; the screen. Does not modify any register
;;; ---------------------------------------------------------
dumpRegs: pushad
;;; put all the registers to print in the stack. The loop
;;; will pop each one, one at a time, once per loop.
push edi
push esi
push edx
push ecx
push ebx
push eax
;;; print top bar
call printBar
;;; prepare to loop 6 times (6 registers )
mov ecx, 6
mov esi, 0
.for pop ebx ;get value of register to convert from stack
push ecx ;save loop counter
lea eax, [regs + esi*2] ;get address of string representing register
mov ax, word [eax] ;ax gets "ax" or "bx" or "cx" or ... "di"
mov word[reg], ax ;modify string with name of register
mov edx, word1 ;edx points to buffer where to store hex digits
;ebx contains register to convert
call storeDWord ;store hex equivalent of ebx in string
mov ecx, string ;print the whole string
mov edx, lineLen
call printInt80
inc esi ;index into register names incremented
pop ecx ;get loop counter back
loop .for
;;; print bottom bar
call printBar
popad
ret
;;; ---------------------------------------------------------
;;; storeDWord: gets
;;; ebx: register to convert
;;; edx: address where to put information
;;; ---------------------------------------------------------
storeDWord: rol ebx, 16 ;bring most sig word in least sig word
call storeWord
inc edx ;make edx skip space in string
rol ebx, 16 ;get lower byte back in least sig pos
call storeWord
ret
;;; ---------------------------------------------------------
;;; storeWord: gets
;;; bx: word to convert
;;; edx: address where to put information. edx incremented by 4
;;; ---------------------------------------------------------
storeWord: rol bx, 8 ;put most sig byte of bx in lower position
call storeByte ;and convert it. Store it in where edx
;points to.
rol bx, 8 ;same with lower nybble
call storeByte
ret
;;; ---------------------------------------------------------
;;; storeByte: gets
;;; bl: byte to convert
;;; edx: address where to store information. edx incremented by 2 automatically
;;; ---------------------------------------------------------
storeByte: push ebx
push eax
rol bl, 4 ;get upper nybble in lower position (LSB)
mov al, bl ;play with al now
and eax, 0x0000000F ;clear all but the least significant nybble
add eax, hex ;get ascii in hex array equivalentn to nybble
mov al, [eax] ;hex digit now in bl
mov byte [edx], al ;store ascii in string
inc edx
rol bl, 4 ;get lower nybble back
mov al, bl ;play with al
and eax, 0x0000000F
add eax, hex
mov al, [eax]
mov byte [edx], al
inc edx
pop eax
pop ebx
ret
;;; ---------------------------------------------------------
;;; printInt80: prints on the screen the string whose address
;;; is in ecx and length in edx.
;;; does not modify registers
;;; ---------------------------------------------------------
printInt80: push eax
push ebx
mov eax, 4
mov ebx, 1
int 0x80
pop ebx
pop eax
ret
;;; ---------------------------------------------------------
;;; printBar: prints the horizontal bar
;;; does not modify registers
;;; ---------------------------------------------------------
printBar: push ecx
push edx
mov ecx, line
mov edx, lineLen
call printInt80
pop edx
pop ecx
ret
Problem #2
;;; ; hw8b.asm
;;; ; Gavi Levy Haskell
;;; ; 231a-ae
;;; ;
;;; ; Uses functions to print the first ten
;;; ; lines of the Pascal Triangle in hex
;;; ;
;;; ; prints:
;;; ;
;;; ; 01 00 00 00 00 00 00 00 00 00
;;; ; 01 01 00 00 00 00 00 00 00 00
;;; ; 01 02 01 00 00 00 00 00 00 00
;;; ; 01 03 03 01 00 00 00 00 00 00
;;; ; 01 04 06 04 01 00 00 00 00 00
;;; ; 01 05 0A 0A 05 01 00 00 00 00
;;; ; 01 06 0F 14 0F 06 01 00 00 00
;;; ; 01 07 15 23 23 15 07 01 00 00
;;; ; 01 08 1C 38 46 38 1C 08 01 00
;;; ; 01 09 24 54 7E 7E 54 24 09 01
;;; ;
;;; ;
;;; ; nasm -f elf -F stabs hw8b.asm
;;; ; ld -melf_i386 -o hw8b hw8b.o
;;; ; ./hw8b
;; -----------------
;; DATA SECTION
;; -----------------
section .data
Pascal times 10 db 0
hexChr db "0123456789ABCDEF"
return db 10
space db " "
;; -----------------
;; CODE SECTION
;; -----------------
section .text
global _start
_start:
mov ebx, Pascal ; pass address of array in ebx
call init ; store 0 in Pascal array and 1
; in first cell
mov ecx, 10
for: mov ebx, Pascal ; pass address of array
call printArray ; print Pascal array
mov ebx, Pascal
call nextLine ; compute next line of triangle
loop for
;;; exit
mov eax, 1
mov ebx, 0
int 0x80
;;; init
;;; recieves address of Pascal in ebx
;;; modifies no registers
init:
mov dword[ebx], 0 ; change first 4 terms to zero
mov dword[ebx + 4], 0 ; change next 4 terms to 0
mov word[ebx + 8], 0 ; change last 2 terms to 0
mov byte[ebx], 1 ; change first term of array to 1
ret
;;; printArray
;;; recieves address of Pascal in ebx
;;; modifies eax, ebx, edx
printArray:
push ecx ; save outer loop position
mov ecx, 10
.for:
push ecx ; save loop position
push ebx ; save address
mov cl, byte[ebx] ; first digit
ror ecx, 4 ; get correct digit
and ecx, 0x0F ; keep only this digit
add ecx, hexChr ; add address of hexChr
mov eax, 4
mov ebx, 1
mov edx, 1
int 0x80 ; print first digit
pop ebx
push ebx
mov cl, byte[ebx] ; second digit
and ecx, 0x0F ; keep only this digit
add ecx, hexChr ; add address of hexChr
mov eax, 4
mov ebx, 1
mov edx, 1
int 0x80 ; print second digit
mov eax, 4
mov ebx, 1
mov ecx, space
mov edx, 1
int 0x80 ; print space
pop ebx
pop ecx ; retrieve loop position
add ebx, 1
loop .for ; retrieve outer loop position
mov eax, 4
mov ebx, 1
mov ecx, return ; return
mov edx, 1
int 0x80 ; next line
pop ecx
ret
;;; nextLine
;;; recieves address of Pascal in ebx
;;; modifies eax, ebx
nextLine:
push ecx ; save outer loop position
mov ecx, 9
add ebx, 9
.for:
push ecx ; save loop position
mov al, byte[ebx - 1] ; n = row[x - 1]
add byte[ebx], al ; row[x] += n
dec ebx ; x -= 1
pop ecx ; retrieve loop position
loop .for
pop ecx ; retrieve loop position
ret
Problem 3: Optional and Extra Credit
Option 1
- First option: write a program that constantly pushes stuff in the stack until it seg-faults. Make it print the size of the stack as it do so.
;;; hwtest.asm
;;; a simple demo program to test I/O with
;;; a C "wrapper" program
%include "asm_io.inc"
;; -------------------------
;; data segment
;; -------------------------
section .data
msg db "Hello! Please enter an integer number: ",0x00
msg2 db "You have entered ",0x00
x dd 0
;; -------------------------
;; code area
;; -------------------------
section .text
global asm_main
asm_main:
mov ebx, esp
for: mov eax, esp
sub eax, ebp
neg eax
call print_int
call print_nl
sub esp, 100000 ;equivalent to pushing 100,000 bytes
jmp for
ret
- The output of the program is:
12
100012
200012
300012
400012
500012
600012
700012
800012
900012
...
11900012
12000012
12100012
12200012
12300012
12400012
12500012
Segmentation fault
- So our program had about 12.5 MBytes of stack when it started. At least that's what was printed when the segmentation fault message was printed, but we cannot be sure that the stack was not actually larger. It is possible that there were many more messages from our program "queued" to come inside the Operating System buffer and ready to be printed when the privileged "Seg Fault" happened.
Option 2
- write an infinite loop program, or program that runs for a long time without using the stack necessarily, and lookup its Process Id. Then look in the /proc directory maintained by Linux how much space is allocated to that process's stack.
Here's our infinite-loop program:
; howMuchStack.asm
; D. Thiebaut
; Infinite loop
;
section .text
global _start
_start:
jmp _start
- Assemble, link, and then run the program
- Figure out what PID the process has. The command for that is ps typed at the prompt:
ps PID TTY TIME CMD 20543 pts/7 00:00:00 tcsh 20641 pts/7 00:00:00 nasmld 20645 pts/7 00:00:03 howMuchStack 20646 pts/7 00:00:00 ps
- Our process Id is shown to be 20645
- Ask Linux what information it has about this process, using the command cat /proc/20645/maps:
cat /proc/20645/maps 08048000-08049000 r-xp 00000000 00:13 9935063 /Users/classes/231a/handout/howMuchStack f7707000-f7708000 r-xp 00000000 00:00 0 [vdso] fff73000-fff94000 rwxp 00000000 00:00 0 [stack]
- subtracting fff73000 from fff94000 we get 0x21000, or 135,168, or 135 KB.
- This is also the information we get by using the command cat /proc/20645/status, whose output is shown below:
cat /proc/20645/status Name: howMuchStack State: R (running) Tgid: 20119 Pid: 20119 PPid: 20060 TracerPid: 0 Uid: 17700 17700 17700 17700 Gid: 17700 17700 17700 17700 FDSize: 64 Groups: 17700 17710 VmPeak: 144 kB VmSize: 144 kB VmLck: 0 kB VmPin: 0 kB VmHWM: 4 kB VmRSS: 4 kB VmData: 4 kB VmStk: 136 kB VmExe: 4 kB VmLib: 0 kB VmPTE: 8 kB VmSwap: 0 kB Threads: 1 SigQ: 0/63717 SigPnd: 0000000000000000 ShdPnd: 0000000000000000 SigBlk: 0000000000000000 SigIgn: 0000000000000000 SigCgt: 0000000000000000 CapInh: 0000000000000000 CapPrm: 0000000000000000 CapEff: 0000000000000000 CapBnd: ffffffffffffffff Cpus_allowed: f Cpus_allowed_list: 0-3 Mems_allowed: 00000000,00000000,00000000,00000000,00000000,... 00000000,00000001 Mems_allowed_list: 0 voluntary_ctxt_switches: 3 nonvoluntary_ctxt_switches: 3470
VmStk is the information for our stack: 136 KB.
Option 3
- Google!
- Here some of the information I was able to find about stack size:
stacks for threads are often smaller. You can change the default at link time, or change at run time also. For reference some defaults are:
o glibc i386, x86_64 7.4 MB
o Tru64 5.1 5.2 MB
o Cygwin 1.8 MB
o Solaris 7..10 1 MB
o MacOS X 10.5 460 KB
o AIX 5 98 KB
o OpenBSD 4.0 64 KB
o HP-UX 11 16 KB
Who controls the default maximum stack size; OS or compiler?
The compiler typically. The OS/hardware does limit it to a certain extent. Default is 8MB on linux IIRC. Think of ulimit -s on Linux (to change stack sizes).
Is the default maximum scaled according to total memory? (ie a machine with 2gb memory would have larger default size than a machine with only 512mb) For this example both machines are same os/compiler setup, just different amounts of system RAM.
No. Until and unless you do it yiurself.You can alter stack sizes via compiler switches.
or
The C++ Standard's take on the issue of stacks and heaps:
The standard is based on an abstract machine and does not really concern itself with hardware or stacks or heaps. It does talk about an allocated store and a free store. The free store is where you'd be if you are calling new (mostly). FWIW, an implementation can have only one memory area masquerading as both stack and heap when it comes to object allocation.
Your question, therefor, boils down to be an implementation specific issue rather than a language issue.
