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# 8086 assembly language

Posts: 3Member
Firstly, let me say i am sorry because my english is bad.i'm not from european or american country. English is not my spoken language in my country.I am electrical power engineering student, programing is not my core subject but i must summit the assignment to complete my study. I need help with this 8086 assembly language, i hope somebody can help me with this question. I have already finish one question about simple calculator . This question is very difficult to me to solved. Could somebody help..please email me at [email protected]

The question is

Write an 8086 assembly language program that will compute:

2.0 * log10 X for x = 0.1, 1.0, 10.0, 100.5, 1000.0, and six other values using a loop.

All values are in the single-precision (short real) format

• Posts: 1,784Member
: Firstly, let me say i am sorry because my english is bad.i'm not from european or american country. English is not my spoken language in my country.I am electrical power engineering student, programing is not my core subject but i must summit the assignment to complete my study. I need help with this 8086 assembly language, i hope somebody can help me with this question. I have already finish one question about simple calculator . This question is very difficult to me to solved. Could somebody help..please email me at [email protected]
:
: The question is
:
: Write an 8086 assembly language program that will compute:
:
: 2.0 * log10 X for x = 0.1, 1.0, 10.0, 100.5, 1000.0, and six other values using a loop.
:
: All values are in the single-precision (short real) format
:
: Thank you for your kindness.
:

I assume you wanna use the FPU.

OK, it's actually pretty simple.

Here's some psuedo(C) of how I would do it:
[code]
for each x in input{
x=2*log10(x)
}
[/code]

Here's a good site for all instructions:
http://www.ousob.com/ng/asm/ng47c.php

But it's missing the FPU instructions...
So I would recomend finding the Intel manuals:
http://developer.intel.com/design/pentium/manuals/

Download one of them (instruction set A-M) and browse for all fpu instructions (all instructions for the fpu starts with an F).

If you wan't to understand how the FPU works(it works a little different than the normal instructions) then this is a good tutorial:
http://www.website.masmforum.com/tutorials/fptute/index.html
• Posts: 3Member
[b][red]This message was edited by amira7x at 2007-3-31 6:55:44[/red][/b][hr]
: : Firstly, let me say i am sorry because my english is bad.i'm not from european or american country. English is not my spoken language in my country.I am electrical power engineering student, programing is not my core subject but i must summit the assignment to complete my study. I need help with this 8086 assembly language, i hope somebody can help me with this question. I have already finish one question about simple calculator . This question is very difficult to me to solved. Could somebody help..please email me at [email protected]
: :
: : The question is
: :
: : Write an 8086 assembly language program that will compute:
: :
: : 2.0 * log10 X for x = 0.1, 1.0, 10.0, 100.5, 1000.0, and six other values using a loop.
: :
: : All values are in the single-precision (short real) format
: :
: : Thank you for your kindness.
: :
:
: I assume you wanna use the FPU.
:
: OK, it's actually pretty simple.
:
: Here's some psuedo(C) of how I would do it:
: [code]
: for each x in input{
: x=2*log10(x)
: }
: [/code]
:
: Here's a good site for all instructions:
: http://www.ousob.com/ng/asm/ng47c.php
:
: But it's missing the FPU instructions...
: So I would recomend finding the Intel manuals:
: http://developer.intel.com/design/pentium/manuals/
:
: Download one of them (instruction set A-M) and browse for all fpu instructions (all instructions for the fpu starts with an F).
:
: If you wan't to understand how the FPU works(it works a little different than the normal instructions) then this is a good tutorial:
: http://www.website.masmforum.com/tutorials/fptute/index.html
:

Thank you friend i appreciate for your help may god bless you.