akol ompen

Could you give me an ecample what is the use of buffer because i cant understand it?

Jeanne Boyarsky wrote:What book are you using that you don't like? This is helpful so you don't get a recommendation that is the same or similar.

stallings

Hello, i want to learn assembly online .My book isnt good its not telling about coding anything.My teachers notes arent good.Can you give me a place that i can learn assembly online books,videos something to help my self.I  working on emu8086.

hello everyone i have a have split a code and trying each part to understand it .I think that i know it but i am not sure about my knowledges so,at this split code i have some comments .Please let me know if i am right or wrong.

1) MOV DS,AX  --->  this for me mean that AX will take the DS.The "variable" lets say

2) LEA DX,MESSAGE ---> LEA means load to DX

3) MOV AH,9 ---->Give(i am not sure if means move) the 9 to AH

4) INT 21H --->stops

5) MOV AL, N   Move N to AL
     MUL AL    --->multiple the AL that means N and N.More specific N*N
6) MOV SI, AX  --->AX goes to SI (i hope with this to remember it good).That means N*N goes to SI?
 
   MOV BL, 2    --->Give 2 to BL
   MOV AL, N    --->Take N and give it to AL
   MUL BL        --->BL * AL?if i am not doing wrong that it is 2* N

7) SUB SI,AX  ---> AX ( N*N) -SI(2*N)
INC SI            ---> ?appear the sub?

8) MOV AX,SI   --->SI goes to AX.2*N goes to AX.Now AX is 2*N
   MOV CL,10  --->10 goes to CL
   DIV CL         -->2*N / 10 ?

9)MOV CL,AH --> AH  "variable" goes to  CL
MOV DL,AL    -->AL    "variable" goes to DL
ADD DL,48     -->DL will be AL +48

akol ompen wrote:

Henry Wong wrote:

akol ompen wrote:
"indentation and spaces surrounding all binary operators" i didnt catch it.If you could do an example please.What i think you mean is that i do lines with out wrote something?

Which is more readable? understandable?

This? ....

for(x=a;x<=b;x=x+h){ while(x<b){ if(abs(x)>1){ y=1/sqrt(pow(x,2)-1); printf("y= %d",y); }else if(abs(x)==1){ y=-9999; printf("y= %d",y); }else if(abs(x)<1){ y=1/sqrt(1-pow(x,2)); printf("y= %d",y); } }                        

Or this? ...

for(x=a;x<=b;x=x+h){    while(x<b){        if(abs(x)>1){            y=1/sqrt(pow(x,2)-1);            printf("y= %d",y);        } else if(abs(x)==1){            y=-9999;            printf("y= %d",y);        } else if(abs(x)<1){            y=1/sqrt(1-pow(x,2));            printf("y= %d",y);        }    } }  

Also, your eariler code had comments. Why did you remove them? Granted, they might not have been the best comments, but you could have improved them, than removing them completely.

Henry

I understand! Thanks a lot now i catch it i will do it
#include <stdio.h> #include <stdlib.h> #include <math.h>   main(){ /*   i give on variables numbers. */ double a=-1.5; double b=1.5; double h=0.1; int x; int y; /* a loop for .We use this variables  x from a until b with "step" h.I think its right the way i wrote it. */ for(x=a;x<=b;x=x+h){   /* the variable x if it is bigger than the b the program will be finished.Thats why i use a while loop */    while(x<b){ /* now if the absolute from  variable x is bigger from 1 then will be a new variable y (in the exercise has f(x)=y=...  , i just dont use the f(x) i hope i am not doing wrong.) . A new variable y that will have the number 1 divide the t root of the second exponential x and this will subtraction the number 1 .More detail 1 up down the root that inculde the exponential power 2 to x and the subraction with 1. I gave it as much as i could more understandable.After will show with the print f the result */           if(abs(x)>1){           y=1/sqrt(pow(x,2)-1);           printf("y= %d",y); /* if the absolute x is equal to 1 then the new variable y will be show the result -9999 and the result will be printed */           }else if(abs(x)==1){            y=-9999;            printf("y= %d",y); /* and the last case if the absolute x is smaller than 1 then it will be y the new variable that will have the 1 divide with the root of 1 substract the x exponential  power 2 /*           }else if(abs(x)<1){           y=1/sqrt(1-pow(x,2));           printf("y= %d",y);           }           } i will do and the next exercise after .Lets finish this.

Henry Wong wrote:

akol ompen wrote:
"indentation and spaces surrounding all binary operators" i didnt catch it.If you could do an example please.What i think you mean is that i do lines with out wrote something?

Which is more readable? understandable?

This? ....

for(x=a;x<=b;x=x+h){ while(x<b){ if(abs(x)>1){ y=1/sqrt(pow(x,2)-1); printf("y= %d",y); }else if(abs(x)==1){ y=-9999; printf("y= %d",y); }else if(abs(x)<1){ y=1/sqrt(1-pow(x,2)); printf("y= %d",y); } }                        

Or this? ...

for(x=a;x<=b;x=x+h){    while(x<b){        if(abs(x)>1){            y=1/sqrt(pow(x,2)-1);            printf("y= %d",y);        } else if(abs(x)==1){            y=-9999;            printf("y= %d",y);        } else if(abs(x)<1){            y=1/sqrt(1-pow(x,2));            printf("y= %d",y);        }    } }  

Also, your eariler code had comments. Why did you remove them? Granted, they might not have been the best comments, but you could have improved them, than removing them completely.

Henry

I understand! Thanks a lot now i catch it i will do it

Campbell Ritchie wrote:Please read about abs(). Who told you to use it? It is not appropriate for the current application.
Please post the original image with the formula; I removed the link for reasons I explained earlier. I can see it, and I think it evaluates to
1.0 ÷ √(|x² − 1.0|)
in which case when x = ±1, the formula will evaluate to ∞.

What is it the formula for?

I haven't looked at the second exercise yet.

"indentation and spaces surrounding all binary operators" i didnt catch it.If you could do an example please.What i think you mean is that i do lines with out wrote something? .Next,I go to line 20 probably you reffer to  fro loop.The variable b is standar i think it doesnt change its b=1.5 the variable x is changing with x=x+h until this will be bigger than the number b.I used pow(x,2) because thats what they told me ,sorry i will fix it.My mistake I will doing it double.I dont know what will happen but its not good as far as i know and from java.Abs() is need on this exercise,why isnt appropriate?i think i should declare #define f(x) 1.0 ÷ √(|x² − 1.0|) but i am not sure about that.I agree 1.0 ÷ √(|x² − 1.0|) this is it  and that x = ±1.The formula is just an exercise to appear the  results of y .
#include <stdio.h> #include <stdlib.h> #include <math.h> main(){ double a=-1.5; double b=1.5; double h=0.1; double x; double y; for(x=a;x<=b;x=x+h){ while(x<b){ if(abs(x)>1){ y=1/sqrt(pow(x,2)-1); printf("y= %d",y); }else if(abs(x)==1){ y=-9999; printf("y= %d",y); }else if(abs(x)<1){ y=1/sqrt(1-pow(x,2)); printf("y= %d",y); } }

Campbell Ritchie wrote:Please tell us the full details; we can only help if we know what is going wrong. What is supposed to happen and what does happen? I think I can see a compiler error somewhere. Add the image to the post, not the link (which I shall delete) because some people are reluctant to click such links. Please explain what the formula is supposed to calculate.

in the second exercise
#include <stdio.h> #include <stdlib.h> #include <math.h> /* its a method .I give name it factorial because thats what it is.I put it outside the main i delclare it there */ int factorial(int n); main(){ /* i give on variables numbers .I give the type int because thats it seems to be .Maybe the sum it could be double */ int x; int n; int sum; int j; int i=1; int myexp=1; /* i am doing a loop for.To count until n.Its a math symbol Sum with S caps lock . Up to this S it has n and down of S it has i=0.On the right of the number x and power i divide this i!.After =  1+ x/1!+x and in power 2 /2!+ x in power 3 /3!+...+x and in power n / n!. Its the type of a guy who started the name with M.Not Taylor series. i cant remember his name */ for(i=1;i<=n;i++){ /* what i am doing next is i  i use the variable sum to save the result of divide the x power to i with the method (which i name it factorial). */ sum=pow(x,i)/ factorial(int i); /* it says to add this on the variable myexp */ my exp=myexp+sum; /* i printf the myexp */ printf("my exp",myexp); } system("Pause"); } /* a method which is factorial .I did what it is on the math type and this i use it to divide with x^i . I dont know what i am doing wrong.This exercise is too difficult for me */ int factorial(int n){ int product=1; for(j=2;j<=n;j++){ product=product*i; return product; } system("Pause");
i dont think is right i think i have done mistakes. On dev-c++ i take this messages.Expected primary -expression before int.j not declared and i was not declared.Its too difficult this exercisee .I dont know what am i doing wrong

Campbell Ritchie wrote:Please tell us the full details; we can only help if we know what is going wrong. What is supposed to happen and what does happen? I think I can see a compiler error somewhere. Add the image to the post, not the link (which I shall delete) because some people are reluctant to click such links. Please explain what the formula is supposed to calculate.

#include <stdio.h> #include <stdlib.h> #include <math.h> main(){ /* i give on variables numbers. */ double a=-1.5; double b=1.5; double h=0.1; int x; int y; /* a loop for .We use this variables  x from a until b with "step" h.I think its right the way i wrote it. */ for(x=a;x<=b;x=x+h){ /* the variable x if it is bigger than the b the program will be finished.Thats why i use a while loop */ while(x<b){ /* now if the absolute from  variable x is bigger from 1 then will be a new variable y (in the exercise has f(x)=y=...  , i just dont use the f(x) i hope i am not doing wrong.) . A new variable y that will have the number 1 divide the t root of the second exponential x and this will subtraction the number 1 .More detail 1 up down the root that inculde the exponential power 2 to x and the subraction with 1. I gave it as much as i could more understandable.After will show with the print f the result */ if(abs(x)>1){ y=1/sqrt(pow(x,2)-1); printf("y= %d",y); /* if the absolute x is equal to 1 then the new variable y will be show the result -9999 and the result will be printed */ }else if(abs(x)==1){ y=-9999; printf("y= %d",y); /* and the last case if the absolute x is smaller than 1 then it will be y the new variable that will have the 1 divide with the root of 1 substract the x exponential  power 2 /* }else if(abs(x)<1){ y=1/sqrt(1-pow(x,2)); printf("y= %d",y); } } i cant edit my code .I did something on it some changes.If i understand good i am not permited to put link (sorry i didnt knew that) and i tryed find how to put image from my pc to the post but i didnt managed to.In any case i will do again if i am somewhere false.Whats its going on this exercise is that it cant be run properly.What i mean is that on  dev-c++ i press up complile & run and it is doing a calculate 0y= that is never finished.What it should do is to show only some numbers.

Morning I did two exercises and in both  i make some mistakes.I dont know what i did wrong.If something is missing or not .

first exercise

I have a = -1.5, b = 1.5, h = 0.1.We use this variables  x from a until b with "step" h (its a for loop ).After i have this [deleted link] and i will  printf the variables x,y.The program  will be finished when x is bigger than b.
what i did :
#include <std.h> #include <stdlib.h> #include <math.h> main(){ int a=-1.5; int b=1.5; double h=0.1; int x; int y; for(x=a;x<=b;x=x+h){ while(x<b){ if(abs(x)>1){ y=1/sqrt(pow(x,2)-1; printf("y= %d",y); else if(abs(x)==1){ y=-9999; printf("y= %d",y); }else if(abs(x)<1){ y=1/sqrt(1-pow(x,2)); printf("y= %d",y); } } system("Pause"); } }

first exercise
we read  x and n( the number of term will be used).We give 1 to the variable i.We give 1 in the variable term.We give the variable 1 to myexp.For each term i=1...n.we calculate each term from previous with the use of reduction type term<--term*x/i.We add the new term on myexp.We printf the x,myexp and exp(x).

what i did:
#include <stdlib.h> #include <math.h> int factorial(int n); main(){ int x; int n; int sum=1; int i=1; int myexp=1; for(i=1;i<=n;i++){ sum=pow(x,i)/ factorial(int i) my exp=myexp+sum; printf("my exp",myexp); } system("Pause"); } int factorial(int n){ int product=1; for(j=1;j<=n;j++){ product=product*i; return product; }

Is there a way to make code unreadable? I mean if you f12button you will see the code html , can somehow hide this? not to be able the visitors to see the code ?

Dave Tolls wrote:If it's the free version:
https://www.ibm.com/developerworks/downloads/im/db2express/

I will go home and I will tell you if I will have problem.Assembly on program emu ,there is a forum here to ask my question that I have on assembly?

It can be hearing "why you can't" or "where you stuck" or "it's easy to download it" I can't download it.Can someone help me ?