CODEWITHSUNDEEP
c
Yura Bihun
Yura Bihun

Reputation: 19

the code does not work error: called object is not a function or function pointerr

#include <math.h>
#include <stdio.h>
 
int main() {
    double x0, x, eps, dx, r;

    x0 = 0.60000;
    x = x0;
    eps = 0.0001;

    do {
        r = x + pow(x, 1 / 2) + pow(x, 1 / 3) - 2.5;
        dx = (r / (1 -  1 / 2 (x , -1 / 2) +  1/3 (x, -1 / 3)));
        x = (x - dx);
    } while ((abs(dx)) > eps);
          
    printf("%f %f\n", x, r);

    return 0;
}

*error: called object is not a function or function pointer * the code does not work, what needs to be done to make everything right

Upvotes: 1

Views: 110

Answers (1)

Gerhardh
Gerhardh

Reputation: 12404

You have tons of syntax and semantic errors:

pow(x, 1 / 2)

Here 1/2 does integer division which means that the resulting 0.5 is chopped to 0. To get floating point results you must ensure that at least one operand is a floating type like 1.0/2.

Then you have a similar piece

1/2(x,-1/2)

Again, due to integer division -1/2 evaluates to 0. Furthermore you are missing some operator before the brackets. That means the compiler expects some function call.

But 2 is not a function. That is probably the place where you get your error message.

If you use , in other situations than a parameter list or variable definition list, it is taken as a "comma operator" that evaluates to the second operand. That means (x,-1/2)evaluates to (-1/2) which again evaluates to 0. In this case I have no idea what you want to achieve with that expression and cannot give some fix for that.

Finally, you use a function that is not suitable for floating point numbers and don't provide a prototype for it as well:

abs(dx)

You should use fabs instead. Usage of abs is also broken because it requires header stdlib.h which you do not include.

Upvotes: 2

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