Recursive function
How does the fRecur function works and what value does it return to main?
Somebody help!!
#include "stdio.h"
enum con{MIN_NUM=0, MAX_NUM=100};
int fRecur(int n);
main(void)
{
int i,suma=0;
for(i=0;i<=MAX_NUM;i++) suma+=i;
printf("El valor de suma es %i\n",suma);
printf("El valor de suma que devuelve fRecur es %i\n",fRecur(MAX_NUM));
return(0);
}
fRecur(int n)
{
if(n==MIN_NUM) return 0;
else return fRecur(n-1)+n;
}
---Programming is like roaming, you never know where you'll end at------
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No cierto, pero - por favor, aqui nos hacemos los Ingles!!
Buena suerte!
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
Actually, from the looks of it, the function will never stop since the number just keeps growing...I think you have waht we call here a "bum algorithm".
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
<fx: runs code>
El valor de suma es 5050
El valor de suma que devuelve fRecur es 5050
Now what was the question?
> How does the fRecur function works
Imagine a call to fRecur(2)
This will result in
return fRecur(2-1)+2;
So now do the same, (2-1) results in fRecur(1)
return fRecur(1-1)+1;
So now do the same, (1-1) results in fRecur(0)
Aha, this is different,
if(n==MIN_NUM) return 0; /* end of the road */
Working your way back fRecur(0) returns 0, so
return fRecur(1-1)+1;
is
return 0+1;
And
return fRecur(2-1)+2;
is
return 1+2;
