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Continued fraction: Difference between revisions
→{{header|BBC BASIC}}: add C solution
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PI = 3.141592653588017
</pre>
=={{header|C}}==
{{works with|ANSI C}}
<lang c>/* calculate approximations for continued fractions */
#include <stdio.h>
/* kind of function that returns a series of coefficients */
typedef double (*coeff_func)(unsigned n);
/* calculates the specified number of expansions of the continued fraction
* described by the coefficient series f_a and f_b */
double calc(coeff_func f_a, coeff_func f_b, unsigned expansions)
{
double a, b, r;
a = b = r = 0.0;
unsigned i;
for (i = expansions; i > 0; i--) {
a = f_a(i);
b = f_b(i);
r = b / (a + r);
}
a = f_a(0);
return a + r;
}
/* series for sqrt(2) */
double sqrt2_a(unsigned n)
{
return n ? 2.0 : 1.0;
}
double sqrt2_b(unsigned n)
{
return 1.0;
}
/* series for the napier constant */
double napier_a(unsigned n)
{
return n ? n : 2.0;
}
double napier_b(unsigned n)
{
return n > 1.0 ? n - 1.0 : 1.0;
}
/* series for pi */
double pi_a(unsigned n)
{
return n ? 6.0 : 3.0;
}
double pi_b(unsigned n)
{
double c = 2.0 * n - 1.0;
return c * c;
}
int main(void)
{
double sqrt2, napier, pi;
sqrt2 = calc(sqrt2_a, sqrt2_b, 1000);
napier = calc(napier_a, napier_b, 1000);
pi = calc(pi_a, pi_b, 1000);
printf("%12.10g\n%12.10g\n%12.10g\n", sqrt2, napier, pi);
return 0;
}</lang>
=={{header|C++}}==
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