Thread: SImpson's Rule

Hey guys, my instructions are to create a program that approximates the area using Simpson's Rule. I set it up the way my book shows, however it is giving me the following error message:
'function' cannot convert from 'double *' to 'double'
and
a warning that says: different types for formal and actual parameter 1

What do these mean?

Here is the part of the code that it applies to:

Code:

double f(double *ptr, double x, int degree)
{
	double sum = 0;
	int i;
	
		for(i = 0; i <= degree; i++)
		{	
			sum = sum + ptr[i] * pow(x,i);
			return(sum);

		}


}

double rectangular(double *ptr, double a, double b, double n,int degree)
{
		double area = 0;
		double totarea = 0;
		double k = 0;
		double width = (b - a) / n;
		double widthplus = (b - a) / n;
		double height = 0;

		for(k = n; k > 0; k--)
		{
				for(degree = degree; degree >= 0; degree--)
				{
					if(degree != 0)
					{
						height = height + (pow(ptr[degree],degree));
					}
					else
					{
						height = height + (ptr[degree]);
					}

				}

				area = height * width;
				totarea = totarea + area;
				widthplus = widthplus + width;

		}
		return(totarea);
	}

double simpsons(double *ptr, double a, double b, double n, int degree)
{
		
	double answer = 0;
	double h;
	double x;
	int i;

		h = (b - a) / n;

		answer = f(ptr,a,degree);
			
			for(i = 1; i <= n; i++)
			{
				x = a + i*h;
				
				answer = answer + 4 * f(ptr,x - (h/2),degree) + 2 * f(ptr,x,degree);
				
			}

			answer = answer - f(ptr,b,degree);

			return(h*answer/6);


			

}

Thanks for the help.

This always looks suspect to me.

Code:

for(i = 0; i <= degree; i++)

How is the array passed to these functions declared? Does it have one "extra" member? [edit]And what is the relation of degree to this array's size?

How.

For example:

Code:

int array[10];

The array may be indexed from 0-9. Therefore the usual idiom is like this.

Code:

for ( i = 0; i < 10; ++i )

Using <= would go one beyond the end of the array.

[edit]And ya know, returning in the middle of the for loop should have been obvious to me earlier, but I've been quite distracted. Should it ever be possible that your loop gets past the first iteration and makes it through the entire loop, you ought to return something.