C++ typedef function in argument

Question

I have a function, that declared next way:

func(int n,double *xk, double (*f)(int x));

I want to assign that into one word like typedef int (*func_t)(int, *double, ...); . I know how to assign function, but do not understand how to describe function in arguments.

Answer 1

You can do

typedef double (*FuncPtr)(int x);  
int func(int n,double *xk, FuncPtr f);

typedef int (*func_t)(int,double*,FuncPtr);

One step at a time to reduce unnecessary confusion and clutter.

Answer 2

typedef int (*func)(int, double*, double (*)(int));

Answer 3

assuming that the function declaration looks like:

int func(int n, double *xk, double (*f)(int x)) {

    // ...
}

then the typedef of a pointer to this function would look like:

typedef int (*funcPointer)(int, double *, int, int, double (*)(int));

// usage:

funcPointer ptrName = func;

assuming you want to give this ptrName to a function called doThis():

int doThis (funcPointer ptrName) {
  return 0;
}

then you can just call it using:

int y = doThis(ptrName);

Best of luck ! =)