C++：如何将函数引用传递给另一个函数？
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C++:How to pass reference-to-function into another function?
提问by Hudson Worden
I have been reading about function pointers and about using them as parameters for other functions. My question is how would you pass a function by reference without using pointers? I have been trying to find the answer on the Internet but I haven't found a good answer. I know that you can pass variables by reference like this: void funct(int& anInt);. How would you do something similar to this but instead of a reference to a variable a reference to a function was the parameter? Also how would you use a reference to the function in a function body?
我一直在阅读有关函数指针以及将它们用作其他函数的参数的信息。我的问题是如何在不使用指针的情况下通过引用传递函数？我一直试图在互联网上找到答案，但我还没有找到好的答案。我知道，你可以通过这样的引用传递变量： void funct(int& anInt);。你会如何做类似的事情，但不是对变量的引用，而是对函数的引用是参数？另外，您将如何在函数体中使用对函数的引用？
回答by Cheers and hth. - Alf
#include <iostream>
using namespace std;
void doCall( void (&f)(int) )
{
f( 42 );
}
void foo( int x )
{
cout << "The answer might be " << x << "." << endl;
}
int main()
{
doCall( foo );
}
Cheers & hth.,
干杯 & hth.,
