Problem on sin(C programming)

fwongmc · #1 09-Dec-2004, 11:41

Dear all,

I am writing a part of function using sin in math,h
I need to use any array to store sinx where x=0 and +0.1 everytime,but the reuslt is failure.I can't get the result,thus I just put in a small program:

CPP / C++ / C Code:

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

void main(){

double input=0.1;
double result;

result=sin(input);

}

The result 1.7*10^(-03),but the answer gives out 0.9833 but not the result,can anyone tell me where I made the mistake?

BobbyMurcerFan · #2 09-Dec-2004, 12:03

I believe it's b/c C/C++ assume the input for trigonometric funcitons is in radians, not degrees. So you have to convert input from degrees to radians before passing it to sin().

HTH.

fwongmc · #3 09-Dec-2004, 12:57

how?I guess what I declear is right

[c]double sin (double)

isn't it?Can anyone give me a hand on it?

davekw7x · #4 09-Dec-2004, 15:15

Quote:

Originally Posted by fwongmc

how?I guess what I declear is right
isn't it?Can anyone give me a hand on it?

Try this: Note the answer:

sin of .1 degrees = 1.745328e-03

sin of .1 radians = 9.983342e-02

CPP / C++ / C Code:

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

int main()
{
  double xrad; /* angle in radians */
  double xdeg; /* angle in degrees */
  double result;

  double pi;

  pi = 4.0 * atan(1.0); /* get pi without having to type in 16 digits */

  xrad = 0.1;
  xdeg = xrad * 180.0 / pi;
  result = sin(xrad);
  printf("For x = %f radians (%f degrees), sin(x) = %e\n", xrad, xdeg, result);

  xdeg = 0.1;
  xrad = xdeg * pi / 180.0;
  result = sin(xrad);
  printf("For x = %f degrees (%f radians), sin(x) = %e\n", xdeg, xrad, result);
  return 0;
}

Regards,

Dave

fwongmc · #5 09-Dec-2004, 22:53

Thanks Dave!

Here comes a question:
I need to calculate sin(x) where x initial is 0.0,and add 0.1 itself after the executoin.The reuslt were store on the array func[]

Here comes my code:

CPP / C++ / C Code:

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

int main ()	{

double func[99];
int choice;
double in_value=0.1;
int num;
double trans=0.0;
double pi;

pi = 4.0 * atan(1.0); /* get pi without having to type in 16 digits */

printf ("Enter function to be plotted:\n");
printf ("For f(x) = sin(x), enter choice 1\nFor f(x) = exp(-0.5x)sin(3x), enter choice 2\n");
scanf ("%d",&choice); 
	if (choice==1)	{	
	 		for (num=0;num<=100;num++) {
				func[num]=sin((trans*pi)/180);
					trans+=in_value;
						printf ("func[%d]=%f\n",num,func[num]);	}
					
			printf ("abcdefg");}
else  
	printf ("You choose 2!\n");
	
}

The result is correct(I have to thanks Dave!)But after Ixecute with Vc++,after execution,a warning message told me that my exe program have problems and ask me to send a report to M$,does my code getting error?(It's a bit urgent!

davekw7x · #6 09-Dec-2004, 23:01

Quote:

Originally Posted by fwongmc

Thanks Dave!

Here comes a question:
I need to calculate sin(x) where x initial is 0.0,and add 0.1 itself after the executoin.The reuslt were store on the array func[]

Here comes my code:

CPP / C++ / C Code:


double func[99];

	 		for (num=0;num<=100;num++) {
				func[num]=sin((trans*pi)/180);
					trans+=in_value;
						printf ("func[%d]=%f\n",num,func[num]);	}

The result is correct(I have to thanks Dave!)But after Ixecute with Vc++,after execution,a warning message told me that my exe program have problems and ask me to send a report to M$,does my code getting error?(It's a bit urgent!

You declare func to be an array with 99 elements. They are func[0], func[1], ... func[98].

Then you try to store something in func[99] and func[100]. That is, the program is running past the end of the array.

Just declare

CPP / C++ / C Code:

double func[101]

Then you can go from func[0] up to and including func[100].

Regards,

Dave

fwongmc · #7 10-Dec-2004, 09:12

Thanks everyone here.I discover that this is an array problem and finally I fix it.
Can anyone tell me how can I correct floating number in double into nearest integer?

I have an array which pass the calculated result to a new array:

CPP / C++ / C Code:

for(i=0;i<=81;i++){
temp[i]=(calculations+0.5)}

what the array temp stores is numbers in double (eg 71.2323232..../69.56566),how can I round it up into(71/69)?I tried to convert the result into the int

CPP / C++ / C Code:

temp[i]=(int)(calculations+0.5);

,but failure.?(it return result on 0.00000000)May I ask only float can change its type by int?,help...!!

davekw7x · #8 10-Dec-2004, 10:23

Quote:

Originally Posted by fwongmc

Thanks everyone here.I discover that this is an array problem and finally I fix it.
Can anyone tell me how can I correct floating number in double into nearest integer?

I have an array which pass the calculated result to a new array:

CPP / C++ / C Code:

for(i=0;i<=81;i++){
temp[i]=(calculations+0.5)}

what the array temp stores is numbers in double (eg 71.2323232..../69.56566),how can I round it up into(71/69)?I tried to convert the result into the int

CPP / C++ / C Code:

temp[i]=(int)(calculations+0.5);

,but failure.?(it return result on 0.00000000)May I ask only float can change its type by int?,help...!!

First note that if you convert a floating point number (float or double) to an int, the fractional part is truncated.

If you want to round to the nearest integer, here is a way

If the number is greater than zero, add 0.5 and convert to an integer
If the number is less than zero, subtract -.5 and convert to an integer.

If you started with a double and you want the result to be a double, just convert the result back to a double.

CPP / C++ / C Code:

  double x;
  int tempint;

  x = 12.56;
  tempint = x + 0.5; /* convert 13.06 to an int */
  x = tempint;       /* convert back to double  */

  x = -12.56        
  tempint = x - 0.5; /* convert -13.06 to an int */
  x = tempint;       /* convert back to a double */

Now, if you want to round to the nearest tenth instead of the nearest integer:

Multiply the original by 10.0
Then round off this number to the nearest int (following the steps above)
Then divide the integer value by 10.0 to get a double that has been rounded to hthe nearest tenth.

CPP / C++ / C Code:

  x = 12.56;
  tempint = x * 10.0 + 0.5; /* convert 126.1 to an int */
  x = tempint / 10.0;        /* convert back to double  */
 
  x = -12.56;
  tempint = x * 10.0 - 0.5; /* convert -126.1 to an int */
  x = tempint / 10.0;        /* convert back to a double */

You don't actually have to store values in a temporary int; you could use casting to perform all operations in a single statement. You could, of course make functions that have these calculations, then you could say something like

CPP / C++ / C Code:

  x = roundint(x);

Where the operations for rounding to the nearest integer are in your function roundint(), and roundint() could look something like

CPP / C++ / C Code:

double roundint(double arg)
{
  if (arg > 0) {
    return (double)((int)(arg + 0.5));
  }
  else{
    return (double)((int)(arg - 0.5));
  }
}

Regards,

Dave

fwongmc · #9 10-Dec-2004, 11:08

thanks for Dave and all of you again and again

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