ARM Compiler and assembly on QEMU

Transcription

1 School of Electrical and Computer Engineering N.T.U.A. Embedded System Design Dimitrios Soudris ARM Compiler and assembly on QEMU

2 Άδεια Χρήσης Το παρόν εκπαιδευτικό υλικό υπόκειται σε άδειες χρήσης Creative Commons. Για εκπαιδευτικό υλικό, όπως εικόνες, που υπόκειται σε άδεια χρήσης άλλου τύπου, αυτή πρέπει να αναφέρεται ρητώς.

3 Still starting... Write and compile a simple hello world.c : gcc -Wall hello_world.c -o hello_world file hello_world Extract assembly: gcc -Wall hello_world.c -S Compile assembly: Gcc -Wall hello_world.s -o hello_world

4 Assembly template.text.global main main: //instructions here.data //few more instructions here

5 Print a number #include <stdio.h> Do it in assembly int main() { int a=15; printf( Number=%d\n,a); }

6 #include <stdio.h> int main() { } int a=15; printf( Number=%d\n,a); Print a number.text.global main.extern printf main:.data push {ip, lr} ldr r0, =string mov r1, #15 bl printf pop {ip, pc} string:.asciz "Number=%d\n"

7 gcd challenge #include <stdio.h> void main() { int a = 53; int b= 43; while (a!= b) { if (a > b) a = a - b; else b = b - a; Compile and run it Check produced assembly Use -O0 and -O3 as compile arguments Can you do it better? Write your own assembly (based on previous file) REMEMBER: Embedded systems is about efficiency } printf( GCD=%d\n,a); }

8 #include <stdio.h> void main() { int a = 53; int b= 43; while (a!= b) { if (a > b) a = a - b; else b = b - a; } printf( GCD=%d\n,a); } gcd challenge.text.global main.extern printf main: mov r3, #53 mov r4, #43 gcd: //your code here push {ip, lr} ldr r0, =string mov r1, r3 bl printf pop {ip, pc}.data string:.asciz "Number=%d\n"

9 gcd challenge: the winner is #include <stdio.h> void main() { int a = 53; int b= 43; while (a!= b) { if (a > b) a = a - b; else b = b - a;.text.global main.extern printf main: mov r3, #53 mov r4, #43 gcd: cmp r3, r4 subgt r3, r3, r4 suble r4, r4, r3 bne gcd push {ip, lr} ldr r0, =string mov r1, r3 } } printf( GCD=%d\n,a); bl printf pop {ip, pc}.data string:.asciz "Number=%d\n"

10 Multiplication Suppose r2=3 and r3=27 Write two assembly programs that implement the arithmetic operation r2*r3 and print the result You can use any command You CANNOT use MUL/MLA commands (tip: think r2=y and r3=27, result=y*27)

11 Multiplication solution.text Suppose r2=3 and r3=27 Write two assembly programs that implement the arithmetic operation r2*r3 and print the result You can use any command You CANNOT use MUL/MLA commands (tip: think r2=y and r3=27, result=y*27).global main.extern printf main: mov r2, #3 mov r3, #27 add r4, r2, r2, LSL #3 add r5, r4, r4, LSL #1 push {ip, lr} ldr r0, =string mov r1, r5 bl printf pop {ip, pc}.data string:.asciz "Number=%d\n"

12 One more If we list all the natural numbers below 10 that are multiples of 3, we get 3, 6 and 9. The sum of these multiples is 18 Find the sum of all the multiples of 3 below 1000

13 Solution If we list all the natural numbers below 10 that are multiples of 3, we get 3, 6 and 9. The sum of these multiples is 18 Find the sum of all the multiples of 3 below 1000.text.global main.extern printf main: loop: mov r2, #0 mov r3, #0 add r2, r2, r3 add r3, r3, #3 cmp r3, #1000 blt loop push {ip, lr} ldr r0, =string mov r1, r2 bl printf pop {ip, pc}.data string:.asciz "Number=%d\n"

14 So far #include <stdio.h> int main() { int y=2014; printf( Hello mlab %d\n,y); }.text.global main.extern printf main: push {ip, lr} ldr r0, =string mov r1, #2014 bl printf pop {ip, pc}.data string:.asciz Hello mlab=%d\n"

15 Linux system calls #include <unistd.h> int open(const char *pathname, int flags); ssize_t read(int fd, void *buf, size_t count); ssize_t write(int fd, const void *buf, size_t count); int close(int fd); void _exit(int status);

16 Unistd.h of arm debian #if defined( thumb ) defined( ARM_EABI ) #define NR_SYSCALL_BASE 0 #else #define NR_SYSCALL_BASE NR_OABI_SYSCALL_BASE #endif #define NR_restart_syscall ( NR_SYSCALL_BASE+ 0) #define NR_exit ( NR_SYSCALL_BASE+ 1) #define NR_fork ( NR_SYSCALL_BASE+ 2) #define NR_read ( NR_SYSCALL_BASE+ 3) #define NR_write ( NR_SYSCALL_BASE+ 4) #define NR_open ( NR_SYSCALL_BASE+ 5) #define NR_close ( NR_SYSCALL_BASE+ 6)

17 System calls - calling convention Man syscalls Syscall number at r7 Arg1 to r0, Arg2 to r1,, Arg7 to r6 Return value to r0 After everything is set, execute swi 0 Software interrupt!

18 Exercise 1 Create a simple program to read a phrase Phrase must be 8 characters long Repeat reading until phrase is precisely 8 chars long Print phrase to output Use a prompt message Stdin has file descriptor 0 Stdout has file descriptor 1 In.data section, use len =. string_name to get the length in bytes of the string (why?) Syscall number at r7 Arg1 to r0, Arg2 to r1,, Arg7 to r6 Return value to r0 After everything is set, execute swi 0 NR_exit -> 1 NR_fork -> 2 NR_read -> 3 NR_write -> 4 NR_open -> 5 NR_close -> 6

19 Exercise 2 Goal is to create a small ciphering program We have an 8 char input string We have an 8 char input pass phrase The i-th encrypted char will result by subtracting the 8-i element of the passphrase Syscall number at r7 Arg1 to r0, Arg2 to r1,, Arg7 to r6 Return value to r0 After everything is set, execute swi 0 NR_exit -> 1 NR_fork -> 2 NR_read -> 3 NR_write -> 4 NR_open -> 5 NR_close -> 6

20 Exercise 3 Create a decryption program Decryption will be the inverse process of exercise 2 Encrypted input read from a file Decrypted output written to a file Passphrase read from screen Decryption must be placed in a subroutine

21 Exercise 3 necessary RD_ONLY = 0 WR_ONLY O_CREAT = 65 Input file name must be a null terminated string! BL command updated link register to enter subroutine Use stmfd to store important registers to stack Sp must be updated (use!) Use multiple registers { r0- r6, fp, lr } (example) Use ldmfd to restore and then branch to main program Syscall number at r7 Arg1 to r0, Arg2 to r1,, Arg7 to r6 Return value to r0 After everything is set, execute swi 0 NR_exit -> 1 NR_fork -> 2 NR_read -> 3 NR_write -> 4 NR_open -> 5 NR_close -> 6

22 Άδεια Χρήσης Το παρόν εκπαιδευτικό υλικό υπόκειται σε άδειες χρήσης Creative Commons. Για εκπαιδευτικό υλικό, όπως εικόνες, που υπόκειται σε άδεια χρήσης άλλου τύπου, αυτή πρέπει να αναφέρεται ρητώς.