Re: more precision in division by C# is possible? - C# Discussion Boards

how to start a messenger project ??

2-Apr-06 12:26

Please don't cross post.

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abdelhameed812-Apr-06 9:59

abdelhameed81

2-Apr-06 9:59

hello all and thank you for your help so far

i would like to start a new project as apractice. it will be a messenger
all what i need to know is how to get started and what are the tools that i have to use

thank you for your concern

abdelhameed81

Re: how to start a messenger project ??

CWIZO2-Apr-06 20:45

CWIZO

2-Apr-06 20:45

Open Visual Studio and click on the New proejct button.

--------------------------------------------------------
My development blog
Q:What does the derived class in C# tell to it's parent?
A:All your base are belong to us!

more precision in division by C# is possible?

luca_la2-Apr-06 9:37

2-Apr-06 9:37

Hi,
i'm writing here because i'm an strange problem.
Taken two points, i' ve to find the step of x of line that goes on two points; to do that i had wrote this function:
// P1(x1,y1) e P2(x2,y2)
double m, dy, dx, ypp, xpp, q;
/*********** F(x) *************/
dy = (double)y2 - (double)y1;
dx = (double)x2 - (double)x1;
m = dy / dx;
q = (double)(x2 * y1 - x1 * y2) / (double)(x2 - x1);
/*********************************************************/
// Point next P1
ypp = (double)(y1 + 1);
xpp = (ypp - q) / m;
return xpp - x1;

Now i have step of x and to go to P2 i do that:
Xstart = x1; Ystart = y1;
for (; Xstart < x2;)
{
Xstart += Passox; Ystart += 1;
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Xstart is double but i need its integer value, so i use a cast in a temporary var.:
Xtmp = (int)Xstart;

but any Xtmp are incorrect.
Now i know that m-value is incorrect, because when i find m by scientific-calculator i had another value: by this value my function work correctly!
Can an scientific calculator do division better tahn C# on Burton processor??
Sorry for my english and thanx for your time,
Luca

Guffa2-Apr-06 12:22

2-Apr-06 12:22

From where do you get the Passox value?

Double calculations are usually correct to 14 decimals. How much incorrect is it?

When you return the value (xpp - x1), is it supposed to be an integer? (Cause it is.)

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luca_la2-Apr-06 14:40

2-Apr-06 14:40

Passox is the value returned: xpp - x1 (Passox translated from italian is step of x).
The function, i use Passox in, receives Passox as double; in this fuction i cast Passox from double to int, but in a temp variable: so i don't change the double var Passox (i do:
Xtmp = (int)Passox ) and so it's no problem of course!
The error for example is this:
start to P1(262, 2), end to P2(772, 309);
my final value is Pend(772, 308.99999999999886) -> after int-cast become Pend(772, 308) but it must be Pend(772, 309)!

I calculated step of y (i had to work with it too) by my scientific calculator (step of x and step of y have the same equation but for step of y i have that example so i report it!); the value is:
m = 1.661237785<-- it's work ok! Pend(772, 309.00000001300782)-> after int-cast become
Pend(772, 309)

value by C# is:
m = 1.6612377850162867<-- more precision but don't work as you can see up! This m by C# > m by
scientific calculator but the y of Pend by C# is less...

I write about only last point of my line but i need precision in all points of line... it's possible? I thinked to do:
PrecisionLost = 309 (y i want) - 308.99999999999886 (y i have);
Passoy = Passoy + PrecisionLost; (or, as the same, Passox = Passox + PrecisionLost)
but if someone knows another solution, i prefer very much! i'm sorry for me but i need maximum precision..
Thanks a lot,
luca

Guffa2-Apr-06 19:56

2-Apr-06 19:56

Your problem is not the precision of the calculations, but the casting of the values.

As you see, the error of the final result is 0.00000001300782 when using the hand calculated value, but only 0.00000000000114 with the value calculated with C#, so the result is at least 10000 times better with the later one.

When you just cast the value to int, you will be rounding it down instead of rounding to the nearest value. Use Math.Round() before casting the value to an int, and you will get the correct value.

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luca_la2-Apr-06 23:21

2-Apr-06 23:21

It's true, it's not precision error thank you! Mathematics doesn't like me!
I tryed to use Math.Round but C# get me this error:

Error 2 Cannot implicitly convert type 'double' to 'int'. An explicit conversion exists (are
you missing a cast?).

If i have to do a cast before using Round(), it's not a solution!
And if i'll can use it, i need to know how much the cost of code increase. Where i can read about?
thanks again,
Luca

Guffa4-Apr-06 2:37

4-Apr-06 2:37

luca_la wrote:
If i have to do a cast before using Round(), it's not a solution!

No, you don't have to do that. The Round method takes a double value. It's after you have done the rounding that you have to do the casting. The method returns a double value, you have to cast that into an integer to store it into an integer variable.

luca_la wrote:
And if i'll can use it, i need to know how much the cost of code increase.

The rounding is done using a single FPU instrution, so it's not expensive at all. It's comparable to casting a double to an integer, which also uses a single FPU instruction.

Consider this line of code:

y = Math.Round(x);

And look the disassembly of the generated code (with my comments):

00000051 fld qword ptr [ebp-44h] ; load from x<br />
00000054 frndint ; round<br />
00000056 fstp qword ptr [ebp-58h] ; store in temp<br />
00000059 fld qword ptr [ebp-58h] ; load from temp<br />
0000005c fstp qword ptr [ebp-4Ch] ; store in y

As you see, the rounding itself is just one instruction, and the rest is just getting the data in and out of the FPU stack and storing the result. If you use the method in an expression, the value doesn't even have to leave the FPU stack between operations.

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luca_la7-Apr-06 3:22