C程序计算x的正弦

Question

This is what i have done so far. I don't know what is wrong with the code. In theory it should run perfectly well (or I might be wrong), but it just doesn't and it is driving me crazy. I'm a beginner BTW.

Can anyone please point out what is wrong with the code?

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

main()
{
   int i , sum = 0 , n;
   float x;
   printf("Please enter the desired values for x and n (n>0): ");
   scanf("%f %d",&x,&n);
   for(i=1;i<=n;i++)
   {
       sum = sum +((pow(-1,i+1)*pow(x,2*i-1))/(factorial(2*i-1)));
   }
   printf("%f",sum);
}

int factorial(int n)
{
   int c;
   int result = 1;

   for( c = 1 ; c <= n ; c++ )
         result = result*c;

   return ( result );
}

Answer 1

Main issues:

• sum type should be float or double , not int ;
• factorial(int n) must be able to return very big numbers, so its return type should be double too.

Possible solution:

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

double factorial(int n)
{
    if (n == 0) return 1;
    return n * factorial(n-1);
}
main()
{
    int n;
    double x, sum = 0;
    printf("Please enter the desired values for x and n (n>0): ");
    scanf("%lf %d", &x, &n);
    for(int i = 0; i <= n; i++)
    {
        sum += pow(-1, i) * pow(x, 2 * i + 1) / (factorial(2 * i + 1));
    }
    printf("%f", sum);
}

To make your sine calculator bulletproof you should add some lines to check the value of the input x and reduce it at least to the domain [-pi, pi] before evaluating the series. Check out my answers here and here to understand why.

Answer 2

@Busy Beaver good answer points out some failings in OP code.

But to dig deeper on how OP could solve this without Stack Overflow .

Can anyone please point out what is wrong with the code?

Rather than seek someone to assist, use your compiler first. Enable all compiler warnings. A good compiler will complain about things like below. This is faster feedback than posting on SO.

// return type defaults to 'int' 
main()
// this should be as below  (amongst other possibilities)
int main(void)

// implicit declaration of function 'factorial'
sum = sum +((pow(-1,i+1)*pow(x,2*i-1))/(factorial(2*i-1)));
// factorial should be declared/defined before it it used

// conversion to 'int' from 'double' may alter its value
sum = sum +((pow(-1,i+1)*pow(x,2*i-1))/(factorial(2*i-1)));
// This is the hint that `sum` should also be a floating point.

// format '%f' expects argument of type 'double', but argument 2 has type 'int'
printf("%f",sum);
// sum is type `int`, the matching specifier is "%d"`.

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By fixing these warnings, code "works" without other changes. Still there are issues of precision, limited range, efficiency and overflow in the factorial() calculation. Lesson to learn: use your compiler to help with the basic issues.

Please enter the desired values for x and n (n>0): 1 5
0.841471

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I now see this is an old post and OP may be have left the building .