- Assembly - Memory Management
- Assembly - File Management
- Assembly - Macros
- Assembly - Recursion
- Assembly - Procedures
- Assembly - Arrays
- Assembly - Strings
- Assembly - Numbers
- Assembly - Loops
- Assembly - Conditions
- Assembly - Logical Instructions
- Assembly - Arithmetic Instructions
- Assembly - Constants
- Assembly - Variables
- Assembly - Addressing Modes
- Assembly - System Calls
- Assembly - Registers
- Assembly - Memory Segments
- Assembly - Basic Syntax
- Assembly - Environment Setup
- Assembly - Introduction
- Assembly - Home
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Assembly - Procedures
Procedures or subroutines are very important in assembly language, as the assembly language programs tend to be large in size. Procedures are identified by a name. Following this name, the body of the procedure is described which performs a well-defined job. End of the procedure is indicated by a return statement.
Syntax
Following is the syntax to define a procedure −
proc_name: procedure body ... ret
The procedure is called from another function by using the CALL instruction. The CALL instruction should have the name of the called procedure as an argument as shown below −
CALL proc_name
The called procedure returns the control to the calpng procedure by using the RET instruction.
Example
Let us write a very simple procedure named sum that adds the variables stored in the ECX and EDX register and returns the sum in the EAX register −
section .text global _start ;must be declared for using gcc _start: ;tell pnker entry point mov ecx, 4 sub ecx, 0 mov edx, 5 sub edx, 0 call sum ;call sum procedure mov [res], eax mov ecx, msg mov edx, len mov ebx,1 ;file descriptor (stdout) mov eax,4 ;system call number (sys_write) int 0x80 ;call kernel mov ecx, res mov edx, 1 mov ebx, 1 ;file descriptor (stdout) mov eax, 4 ;system call number (sys_write) int 0x80 ;call kernel mov eax,1 ;system call number (sys_exit) int 0x80 ;call kernel sum: mov eax, ecx add eax, edx add eax, 0 ret section .data msg db "The sum is:", 0xA,0xD len equ $- msg segment .bss res resb 1
When the above code is compiled and executed, it produces the following result −
The sum is: 9
Stacks Data Structure
A stack is an array-pke data structure in the memory in which data can be stored and removed from a location called the top of the stack. The data that needs to be stored is pushed into the stack and data to be retrieved is popped out from the stack. Stack is a LIFO data structure, i.e., the data stored first is retrieved last.
Assembly language provides two instructions for stack operations: PUSH and POP. These instructions have syntaxes pke −
PUSH operand POP address/register
The memory space reserved in the stack segment is used for implementing stack. The registers SS and ESP (or SP) are used for implementing the stack. The top of the stack, which points to the last data item inserted into the stack is pointed to by the SS:ESP register, where the SS register points to the beginning of the stack segment and the SP (or ESP) gives the offset into the stack segment.
The stack implementation has the following characteristics −
Only words or doublewords could be saved into the stack, not a byte.
The stack grows in the reverse direction, i.e., toward the lower memory address
The top of the stack points to the last item inserted in the stack; it points to the lower byte of the last word inserted.
As we discussed about storing the values of the registers in the stack before using them for some use; it can be done in following way −
; Save the AX and BX registers in the stack PUSH AX PUSH BX ; Use the registers for other purpose MOV AX, VALUE1 MOV BX, VALUE2 ... MOV VALUE1, AX MOV VALUE2, BX ; Restore the original values POP BX POP AX
Example
The following program displays the entire ASCII character set. The main program calls a procedure named display, which displays the ASCII character set.
section .text global _start ;must be declared for using gcc _start: ;tell pnker entry point call display mov eax,1 ;system call number (sys_exit) int 0x80 ;call kernel display: mov ecx, 256 next: push ecx mov eax, 4 mov ebx, 1 mov ecx, achar mov edx, 1 int 80h pop ecx mov dx, [achar] cmp byte [achar], 0dh inc byte [achar] loop next ret section .data achar db 0
When the above code is compiled and executed, it produces the following result −
0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|} ... ...Advertisements