Bezier curve.

warny_maelstrom · #1 04-Mar-2004, 03:51

My code has no errors or warnings but it just closes after asking for the values for w, and i have no clue y. Im trying to work out the coordinates on a bezier curve, the maths is fine i think but the code just doesnt work like i want it to.

CPP / C++ / C Code:

#include<stdio.h>
#include<math.h>

main()
{
float u,g,p,s,t,x,y;
int xr0,xr1,xr2,xr3,yr0,yr1,yr2,yr3,w0,w1,w2,w3;

u=0;


printf("Enter in r values for the x coordinates e.g. x0,x1,x2,x3\n");
scanf("%d,%d,%d,%d",&xr0,&xr1,&xr2,&xr3);
printf("Enter in r values for the y coordinates e.g. y0,y1,y2,y3\n");
scanf("%d,%d,%d,%d",&yr0,&yr1,&yr2,&yr3);
printf("Enter in w values e.g. w0,w1,w2,w3\n");
scanf("%d,%d,%d,%d",&w0,&w1,&w2,&w3);


if (u=0,u<1)
        {
        p=(pow(1-u,3));
        g=(3*(pow(1-u,2)*u));
        s=(3*pow((1-u)*u,2));
        t=pow(u,3);

        x=((p*w0*xr0)+(g*w1*xr1)+(s*w2*xr2)+(t*w3*xr3))/((p*w0)+(g*w1)+(s*w2)+(t*w3));
        printf("\n",x);

        y=((p*w0*yr0)+(g*w1*yr1)+(s*w2*yr2)+(t*w3*yr3))/((p*w0)+(g*w1)+(s*w2)+(t*w3));
        printf("\t",y);
        u=(u+0.2);
        }
else
return 0;

}

Edit: i managed to solve the first problem myself so now the title is wrong and misleading but i couldnt delete this post.

dsmith · #2 04-Mar-2004, 08:42

Warny, you may want to repost your code if you fixed the problem and are still having problems with something else. I see several things that look wrong in this program, but I don't want to comment on something that you already fixed.

WaltP · #3 04-Mar-2004, 09:15

And be specific. "It doesn't do what I want it to" is not a problem. What it does do is, as well as what it's supposed to do. Both these are needed for a solution.

davekw7x · #4 04-Mar-2004, 09:23

Here's a suggestion whenever your program doesn't give the answers that
you expect:

Every time you obtain input values (from user console input or from a
file), print out their values. For example:

CPP / C++ / C Code:

printf("Enter in r values for the x coordinates e.g. x0,x1,x2,x3\n");
scanf("%d,%d,%d,%d",&xr0,&xr1,&xr2,&xr3);
printf("Debug: Here is what you entered: <%d> <%d> <%d> <%d>\n\n", xr0, xr1, xr2, xr3);

Note that using scanf() requires the user to input the data exactly as indicated
in the specificaton. So, for example, the user must enter integer values
separated by commas in each case.

I am sure that your if( ) condition is not doing what you had in mind,
but: first things first. Verify the input!

Dave

warny_maelstrom · #5 04-Mar-2004, 10:52

My code works perfectly now i managed to fix my problem, which as i didnt state it b4 was the fact that i wasnt given any output for values of x and y. Now when i print the values of x and y i get the output:
x1 y5
x2 y6
x3 y7
whereas id prefer the numbers for the y values to b y1,y2,y3. Is there any way of doing this.

CPP / C++ / C Code:

#include<stdio.h>
#include<math.h>

int main(void) /* main returns an int */
{
    float u, g, p, s, t, x, y;
    int xr0, xr1, xr2, xr3, yr0, yr1, yr2, yr3, w0, w1, w2, w3;
    u = 0;

    printf("Enter in r values for the x coordinates e.g. x0,x1,x2,x3\n");
    scanf("%d,%d,%d,%d", &xr0, &xr1, &xr2, &xr3);
    printf("Enter in r values for the y coordinates e.g. y0,y1,y2,y3\n");
    scanf("%d,%d,%d,%d", &yr0, &yr1, &yr2, &yr3);
    printf("Enter in w values e.g. w0,w1,w2,w3\n");
    scanf("%d,%d,%d,%d", &w0, &w1, &w2, &w3);

for ( u = 0 ; u < 1 ; u += 0.2 ) {
        p = (pow(1 - u, 3));
        g = (3 * (pow(1 - u, 2) * u));
        s = (3 * pow((1 - u) * u, 2));
        t = pow(u, 3);

        x = ((p * w0 * xr0) + (g * w1 * xr1) + (s * w2 * xr2) + (t * w3 * xr3)) / ((p * w0) + (g * w1) + (s * w2) + (t * w3));
        printf("\nx%f", x); 

        y = ((p * w0 * yr0) + (g * w1 * yr1) + (s * w2 * yr2) + (t * w3 * yr3)) / ((p * w0) + (g * w1) + (s * w2) + (t * w3));
        printf("\ty%f\n\n", y); 
        u = (u + 0.2);
        }
    return 0;
}

dsmith · #6 04-Mar-2004, 11:30

Quote:

Originally Posted by warny_maelstrom

whereas id prefer the numbers for the y values to b y1,y2,y3. Is there any way of doing this.

I don't know what you mean here. Do you mean that you want the values of y to be 1,2,3 respectively as you go through your loop? Wouldn't that depend on your input values of yr0 through yr3?

Also, did you mean to double index your for loop? (you are incrementing by 0.4 each time you go through the loop

CPP / C++ / C Code:



for ( u = 0 ; u < 1 ; u += 0.2 ) {
         //....
        u = (u + 0.2);
        }

davekw7x · #7 04-Mar-2004, 11:37

Well, I claim that if the output doesn't look like you want it to, that the "code" isn't working "perfectly". (Just my little attempt at humor; but, seriously, output is an important part of the code --- actually it is (usually) the only thing that the user sees.)

You want to print the results for what value(s) of u?

As a debugging aid, I suggest the following:

In place of your statement

CPP / C++ / C Code:

printf("\nx%f", x);

I suggest you put

CPP / C++ / C Code:

printf("For u = %f:   ", u);
printf("\nx = %f", x);

and see if it's really what you want.

Similarly, for the y value, try

CPP / C++ / C Code:

printf("\ty = %f\n\n", y);

Tell us how it turns out.

Regards,

Dave

warny_maelstrom · #8 04-Mar-2004, 12:25

Quote:

Originally Posted by dsmith

Also, did you mean to double index your for loop? (you are incrementing by 0.4 each time you go through the loop

k, screw wot i was asking b4 coz uv alrdy solved it or i realised that it was a really dumb question(hopeing u dont realise wot im refering to). Some im all good now cheers for helping, i might need more help

if i realise that i want more from the program.

WaltP · #9 04-Mar-2004, 13:12

Quote:

Originally Posted by warny_maelstrom

k, screw wot i was asking b4 coz uv alrdy solved it...

I don't know about the rest of you, but I'd prefer posts are written in English as opposed to lazy-geek-ish. English will be required when you get into the business sector for communication, so you might as well learn it now. Also, my brain won't hurh trying to translate messages.

aaroncohn · #10 04-Mar-2004, 13:19

Yeah, honestly, I think it's appalling how many people use such bad grammar and spelling when communicating over the internet. It's still communication, and the person on the other side is still going to get the idea that you can't speak english.

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