Program for declare the threshold, Assembly Language

1. Start your program at address $8500. To do this you need to inform the assembler, through the EQU and ORG assembler directives, that you want your program to start at $8500. This can be done as shown below:

 (leave 2 tab-spaces) ROMSTART EQU $8500

 ORG ROMSTART

The first line above equates the string ROMSTART with $8500. If you mention ROMSTART in any subsequent part of your program, it will be replaced with $8500 during execution.

The second line tells the assembler that the assembly process should now proceed from the address specified by ROMSTART. (ORG can be used multiple times in your program if you wish to alter the assembly flow).

2. Read a set of three hexadecimal numbers located in memory addresses $4000, $4001, and  $4002 and store them to memory locations $0800, $0801, and $0802, respectively.

3. Read the three values from the new memory locations and check each one against a threshold value of $55. You can use the EQU directive at the beginning of the program to declare the threshold as follows:

THRESH EQU $55

4. If the value stored in memory location $0800 is higher (unsigned numbers) than the threshold value, compute the sum of the first five natural numbers. Store the number back to the memory location $0800.

5. If the value stored in memory location $0801 is higher (unsigned numbers) than the threshold value, store $00 to the same memory location.

6. If the value stored in memory location $0802 is higher (unsigned numbers) than the threshold value, subtract $10 from the value stored in the memory location $0802 and store the resulting value back to the same memory location.

Posted Date: 3/8/2013 12:08:57 AM | Location : United States







Related Discussions:- Program for declare the threshold, Assignment Help, Ask Question on Program for declare the threshold, Get Answer, Expert's Help, Program for declare the threshold Discussions

Write discussion on Program for declare the threshold
Your posts are moderated
Related Questions

Interrupt System Based on Multiple 8259As A multiple 8259A interrupt system is diagrammed in given figure in this figure data bus drivers are not indicated, but they could be i

Interrupt System Based on Single 8259 A The 8259A is contained in a 28-pin dual-in-line package that need only a + 5-V supply voltage.  Its organization is shown in given figur


SHORT  : The  SHORT operator denoted to the assembler that only one byte is needed to code the displacement for a jump (for example displacement is within -128 to +127 bytes fr

Cache components The cache sub-system may be divided into 3 functional blocks: Tag RAM, SRAM and theCache Controller. In real designs, these blocks can be implemented  by multi

ROL : Rotate Left without Carry: This instruction rotates the content of the destination operand to the left by the specified count bit-wise excluding the carry. The most signific

I am running a small minecraft server off of my old mac mini, and am having a big issue. My computer isn''t very good, and even just running this server is an issue. I use a comma

Open notepad and enter the code for a program that calculates the following arithmetic expression: x = a + b + c - d - e + f The operands a, b, c, d, e, f, and x should be declared

how o create the flow chart for scan ROW4, Column 1 and 3.tq