C ++将函数的参数传递给另一个函数

Question

I am trying to pass function to a function, declaration looks like

double fun1(const double* ang, std::function<double(double)> f1)

In the main() , I am creating function pointer and pointing it to another function as

std::function<double(double)> f2 = &fun5; //line1

and calling function fun1 as

double x = fun1(&a,f2);

This works. But, if I call function as

double x = fun1(&a,&f2);

,it gives error as

could not convert '& f2' from 'std::function*' to 'std::function

Creating reference works at line1 but calling function with reference does not work. Any idea why is this happening?

Answer 1

• fun5 is a function
• &fun5 is a pointer to a function of the type: double (*)(double)
• f2 is a functor object specifically of the type: function<double(double)>
• &f2 is the address of this functor object it has the type: function<double(double)>*

Because fun1 expects a functor object passing it a pointer to the functor object is illegal. Think about this in terms of int s if that's simpler. If for example I have a function: void foo(int arg) , and I have an int bar I cannot do:

 
 
 
  
  foo(&bar)
 
  

This whould simply be passing an int* as arg , which is obviously illegal. What would have to happen is:

foo(bar)

Similarly your function wants a functor object. You need to pass it the functor object not the pointer, so:

fun1(&a, f2)

Answer 2

首先让我们了解情况，首先您创建了f2，它是＆fun5，所以基本上您已经创建了fun5的别名，它是f2，现在根据您的函数参数，第二个参数需要一个函数f2，您在第一种情况下已通过但是在第二种情况下，您需要有一个称为指针的特殊变量，以指向您已传递的引用/地址，因此将第二个参数修改为* something，第二种情况就可以了。