作为模板参数传递的C ++函数与参数 [英] C++ Function passed as Template Argument vs Parameter

问题描述

在C ++中，有两种方法可以将函数传递给另一个看起来相当的函数。

In C++, there are two ways of passing a function into another function that seem equivalent.

#include <iostream>

int add1(int i){ return i+1; }
int add2(int i){ return i+2; }

template <int (*T)(int) >
void doTemplate(int i){
    std::cout << "Do Template: " << T(i) << "\n";
}

void doParam(int i, int (*f)(int)){
    std::cout << "Do Param: " << f(i) << "\n";
}

int main(){
    doTemplate<add1>(0);
    doTemplate<add2>(0);

    doParam(0, add1);
    doParam(0, add2);
}

doTemplate接受函数作为模板参数，而doParam接受函数

doTemplate accepts a function as a template argument, whereas doParam accepts it as a function pointer, and they both seem to give the same result.

使用每个方法之间的折衷是什么？

What are the trade-offs between using each method?

推荐答案

基于模板的版本允许编译器内联调用，因为函数的地址在编译时是已知的。显然，缺点是函数的地址在编译时是已知的（因为你使用它作为模板参数），有时这可能是不可能的。

The template-based version allows the compiler to inline the call, because the address of the function is known at compile-time. Obviously, the disadvantage is that the address of the function has to be known at compile-time (since you are using it as a template argument), and sometimes this may not be possible.

这就带来了第二种情况，其中函数指针只能在运行时确定，因此编译器不可能执行内联，但是给你灵活性在运行时确定要调用的函数：

That brings us to the second case, where the function pointer may be determined only at run-time, thus making it impossible for the compiler to perform the inlining, but giving you the flexibility of determining at run-time the function to be called:

bool runtimeBooleanExpr = /* ... */;
doParam(0, runtimeBooleanExpr ? add1 : add2);

但请注意，还有第三种方式：

Notice, however, that there is a third way:

template<typename F>
void doParam(int i, F f){
    std::cout << "Do Param: " << f(i) << "\n";
}

这给你更多的灵活性，并且仍然具有在编译时知道的优势什么函数被调用：

Which gives you more flexibility and still has the advantage of knowing at compile-time what function is going to be called:

doParam(0, add1);
doParam(0, add2);

还允许传递任何可调用对象而不是函数指针：

And it also allows passing any callable object instead of a function pointer:

doParam(0, my_functor());

int fortyTwo = 42;
doParam(0, [=] (int i) { return i + fortyTwo; /* or whatever... */ }

为了完整起见，还有一种第四种方法，使用 std :: function ：

For completeness, there is also a fourth way, using std::function:

void doParam(int x, std::function<int(int)> f);

这具有相同的通用性（可以传递任何可调用对象），但也允许您在运行时确定可调用对象 - 最可能的性能损失，因为（再一次）内联对于编译器是不可能的。

Which has the same level of generality (in that you can pass any callable object), but also allows you to determine the callable object at run-time - most likely with a performance penalty, since (once again) inlining becomes impossible for the compiler.

有关最后两个选项的进一步讨论，还见 此问题在StackOverflow上的问答 。

For a further discussion of the last two options, also see this Q&A on StackOverflow.

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