"nan" output error

[Oliveira]

Hi mates, I'm having problems with a "nan" output
Look the code below

#include <stdio.h>
#include <stdlib.h>

float raiz(float n, float *p);

int main()
{
    float n, r = 1, *p;
    p = &r;

    do
    {

        printf("Entre com o número cuja raiz será calculada\n");
        scanf("%f", &n);

        if(n < 0)
            printf("Não há raiz quadrada real para números negativos\n");
    }while(n < 0);

    printf("O valor da raiz quadrada de %.2f é %.2f\n", n, raiz(n, p));

    return 0;
}

float raiz(float n, float *p)
{
    if(abs(*p * *p - n) < 0.001){
	printf("%.2f", *p);
        return (*p);
    }
    else{
        *p = (*p * *p + n)/(2 * *p);
        raiz(n, p);
    }
}

I think I'm not doing well the return of the function "raiz", because the value is well shown inside the function...

Thanks in advance.

[Sebastiani]

I haven't really looked at the algorithm much but one thing I noticed is that you're not returning a value from the recursive call.

[Sebastiani]

One more thing: abs returns an int, so the condition will only be true if it returns 0, obviously.

[Oliveira]

Thank you very much mate.

I've fixed the code and now works fine:

#include <stdio.h>
#include <stdlib.h>

int raiz(float n, float *p);

int main()
{
    float n, r = 1, *p;
    p = &r;

    do
    {

        printf("Entre com o número cuja raiz será calculada\n");
        scanf("%f", &n);

        if(n < 0)
            printf("Não há raiz quadrada real para números negativos\n");
    }while(n < 0);

	raiz(n, p);

    printf("O valor da raiz quadrada de %.2f é %.2f\n", n, r);

    return 0;
}

int raiz(float n, float *p)
{
    if(abs(*p * *p - n) < 0.001){
        return 0;
    }
    else{
        *p = (*p * *p + n)/(2 * *p);
        raiz(n, p);
    }
}

hey, the problem with the previous code was the abs function? Or I did bad the returning?

Thanks.

[BEN10]

NaN(not a number) error generally occurs if the ouput is indeterminate i.e whose value couldn't be determined such as 0/0.

[nonoob]

Probably should use fabs().

Your second post of code looks the same as the first.

[brewbuck]

There is no need to take the absolute value. This is Newton's algorithm. The approximations will converge monotonically to the true value of the square root. So if the initial guess was less than the true root (which will be the case for any input greater than 1, since you start with an initial approximation of 1), then the subsequent approximations will also be less than the true root.

Replace:

fabs( *p * *p - n )

With:

n - *p * *p