将函数用作函数的参数时出错

Question

I'm trying to create a program to numerically integrate a function between two limits. I've created a minimal example (where the "integral" is always 1.0) to illustrate an error I get. The code below tries to use a function whose arguments are two doubles and a function from doubles to doubles:

#include <iostream>
#include <math.h>
double cons(double a, double b, double c(double d))
{        
    return 1.0;
}
double f(double x)
{
    return x*x;
}
int main()
{
    double x;
    double I = cons(0, 1, f(x));
    std::cout << I << "";
}

This results in an error on cpp.sh :

14:31: error: cannot convert 'double' to 'double ( )(double)' for argument '3' to 'double cons(double, double, double ( )(double))'

Obviously, the difference between doubles and double-valued functions is causing a problem here. How can I get this working?

Answer 1

You need to pass the function, not call it.

#include <iostream>
#include <math.h>
double cons(double a, double b, double c(double d))
{        
    return 1.0;
}
double f(double x)
{
    return x*x;
}
int main()
{
    double x;
    double I = cons(0, 1, f);
    std::cout << I << "";
}

Answer 2

You didn't pass a function but the result of a function, a double. Second, you didn't correctly declared a function pointer as argument.

If you want to pass a double then declare a double as argument:

double cons(double a, double b, double c)
{        
    return 1.0*a*b*c;
}
double f(double x)
{
    return x*x;
}
int main()
{
    double x;
    double I = cons(0, 1, f(x));
    std::cout << I << "";
}

If you want to pass a function (aka function pointer as C++ is not a functional language):

double cons(double a, double b, double (*c)(double d))
{        
    return 1.0*c(a);
}
double f(double x)
{
    return x*x;
}
int main()
{
    double x;
    double I = cons(0, 1, f);
    std::cout << I << "";
}

Answer 3

Your problem in this case is that you are calling the function with:

double I = cons(0, 1, f(x));

so you actually give cons the return value of f() instead of the actual function. So you need to write this insteed:

double I = cons(0, 1, f);

As an alternativ you could also use lambda expressions for your problem. For example something like this:

#include <iostream>
#include <math.h>
#include <functional>
double cons(double a, double b, const std::function<double(double)>& callback)
{        
    // do something 
}

int main()
{
    double x;
    double I = cons(0, 1, [](double x){ return x*x});
    std::cout << I << "";
}