Assembly - Arrays

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We have already discussed that the data definition directives to the assembler are used for allocating storage for variables. The variable could also be initialized with some specific value. The initialized value could be specified in hexadecimal, decimal or binary form.

For example, we can define a word variable 'months' in either of the following way −

MONTHS	DW	12
MONTHS	DW	0CH
MONTHS	DW	0110B

The data definition directives can also be used for defining a one-dimensional array. Let us define a one-dimensional array of numbers.

NUMBERS	DW  34,  45,  56,  67,  75, 89

The above definition declares an array of six words each initialized with the numbers 34, 45, 56, 67, 75, 89. This allocates 2x6 = 12 bytes of consecutive memory space. The symbolic address of the first number will be NUMBERS and that of the second number will be NUMBERS + 2 and so on.

Let us take up another example. You can define an array named inventory of size 8, and initialize all the values with zero, as −

INVENTORY   DW  0
            DW  0
            DW  0
            DW  0
            DW  0
            DW  0
            DW  0
            DW  0

Which can be abbreviated as −

INVENTORY   DW  0, 0 , 0 , 0 , 0 , 0 , 0 , 0

The TIMES directive can also be used for multiple initializations to the same value. Using TIMES, the INVENTORY array can be defined as:

INVENTORY TIMES 8 DW 0

Example

The following example demonstrates the above concepts by defining a 3-element array x, which stores three values: 2, 3 and 4. It adds the values in the array and displays the sum 9 −

Live Demo 
section	.text
   global _start   ;must be declared for linker (ld)
	
_start:	
 		
   mov  eax,3      ;number bytes to be summed 
   mov  ebx,0      ;EBX will store the sum
   mov  ecx, x     ;ECX will point to the current element to be summed

top:  add  ebx, [ecx]

   add  ecx,1      ;move pointer to next element
   dec  eax        ;decrement counter
   jnz  top        ;if counter not 0, then loop again

done: 

   add   ebx, '0'
   mov  [sum], ebx ;done, store result in "sum"

display:

   mov  edx,1      ;message length
   mov  ecx, sum   ;message to write
   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

section	.data
global x
x:    
   db  2
   db  4
   db  3

sum: 
   db  0

When the above code is compiled and executed, it produces the following result −

9
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