函数指针作为参数的C ++问题

Question

I am trying to apply newtons method in C++ and right now just testing out if my pointers work and are correct. Now the issue is it cannot call the function to test this, it says there is an issue with converting.

My code:

#include <iostream>
#include <cstdlib>
#include <cmath>

using namespace std;

double newton(double (*f) (double), double (*fPrime)(double), double intialValue, int    iterations);
double f(double x);
double fPrime(double x);

int main() {


int limitIterations = 0;
double intialValue = 0;


cout << "Please enter a starting value for F(X): " ;
cin >> intialValue;

cout << endl << "Please enter the limit of iterations performed: " ;
 cin >> limitIterations;



cout << newton(intialValue, limitIterations);


return 0;
}

 double f(double x) {
 double y;
 y =(x*x*x)+(x*x)+(x);
 return (y);
 }

double fPrime(double x){
double y;
y = 3*(x*x) + 2 * x + 1;
return (y);

}

double newton(double (*f) (double), double (*fPrime)(double), double intialValue, int     iterations){

    double approxValue = 0;

    approxValue = f(intialValue);

    return (approxValue);

 }

And the error:

|26|error: cannot convert 'double' to 'double (*)(double)' for argument '1' to 'double  newton(double (*)(double), double (*)(double), double, int)'|

Answer 1

You don't need to pass function pointers. The functions that you defined above can be used directly. Just define your newton function like this:

double newton(double intialValue, int iterations) {
    double approxValue = 0;
    approxValue = f(intialValue);
    return (approxValue);
}

Answer 2

If you do want to declare newton to take function pointers, then you will need to pass them in at the call-site to newton :

cout << newton(f, fPrime, initialValue, iterations);

The error from the compiler is simply saying that you passed a double in a slot where it was expecting a function pointer and that it has no clue how to convert a double into a function pointer double (*)(double) .