C++ 如何将函数存储到变量？

How do I store a function to a variable?

提问by NullVoxPopuli

I think they are called functors? (it's been a while)

我认为他们被称为函子？（有一阵子了）

Basically, I want to store a pointer to a function in a variable, so I can specify what function I want to use from the command line.

基本上，我想在一个变量中存储一个指向函数的指针，这样我就可以从命令行指定我想使用的函数。

all the functions return and take the same values.

所有函数都返回并取相同的值。

unsigned int func_1 (unsigned int var1)
unsigned int func_2 (unsigned int var1)

function_pointer = either of the above?

so then I could call it by going: function_pointer(my_variable)?

那么我可以通过以下方式调用它： function_pointer(my_variable)?

EDIT: as per @larsmans's suggestion, I've gotten this: Config.h:

编辑：根据@larsmans 的建议，我得到了这个：Config.h：

class Config
{
public:
    unsigned static int (*current_hash_function)(unsigned int);
};

Config.cpp:

配置.cpp：

#include "Config.h"
#include "hashes.h"
unsigned static int (*current_hash_function)(unsigned int) = kennys_hash_16;

hashes.h:

哈希值.h：

unsigned int kennys_hash(unsigned int out);
unsigned int kennys_hash_16(unsigned int out);

hashes.cpp:

hashes.cpp：

just implements the functions in the header

main.cpp:

主.cpp：

#include "Config.h"
#include "hashes.h"
// in test_network:
    unsigned int hashed = Config::current_hash_function(output_binary);

//in main():
        else if (strcmp(argv[i], "-kennys_hash_16") == 0)
        {
            Config::current_hash_function = kennys_hash_16;
        }
        else if (strcmp(argv[i], "-kennys_hash_8") == 0)
        {
            Config::current_hash_function = kennys_hash;
        }

the error I get:

我得到的错误：

g++ -o hPif src/main.o src/fann_utils.o src/hashes.o src/Config.o -lfann -L/usr/local/lib 
Undefined symbols:
  "Config::current_hash_function", referenced from:
      test_network()     in main.o // the place in the code I've selected to show
      auto_test_network_with_random_data(unsigned int, unsigned int, unsigned int)in main.o
      generate_data(unsigned int, unsigned int, unsigned int)in main.o
      _main in main.o // the place in the code I've selected to show
      _main in main.o // the place in the code I've selected to show
      generate_train_file()     in fann_utils.o
ld: symbol(s) not found
collect2: ld returned 1 exit status
make: *** [hPif] Error 1

回答by Jon

The simplest you can do is

你能做的最简单的是

unsigned int (*pFunc)(unsigned int) = func_1;

This is a bare function pointer, which cannot be used to point to anything other than a free function.

这是一个裸函数指针，不能用于指向除自由函数以外的任何内容。

You can make it less painful if your compiler supports the C++0x autokeyword:

如果您的编译器支持 C++0xauto关键字，您可以减少痛苦：

auto pFunc = func_1;

In any case, you can call the function with

在任何情况下，您都可以调用该函数

unsigned int result = pFunc(100);

There are many other options that provide generality, for example:

还有许多其他选项可以提供通用性，例如：

• You can use boost::functionwith any C++ compiler
• With a compiler implementing features of C++0x you can use std::function

• 您可以boost::function与任何 C++ 编译器一起使用
• 使用实现 C++0x 功能的编译器，您可以使用 std::function

These can be used to point to any entity that can be invoked with the appropriate signature (it's actually objects that implement an operator()that are called functors).

这些可以用来指向任何可以用适当的签名调用的实体（它实际上是实现了operator()被称为函子的对象）。

Update (to address updated question)

更新（解决更新的问题）

Your immediate problem is that you attempt to use Config::current_hash_function(which you declare just fine) but fail to define it.

您的直接问题是您尝试使用Config::current_hash_function（您声明很好）但未能定义它。

This defines a global static pointer to a function, unrelated to anything in class Config:

这定义了一个指向函数的全局静态指针，与 中的任何内容无关class Config：

unsigned static int (*current_hash_function)(unsigned int) = kennys_hash_16;

This is what you need instead:

这是你需要的：

unsigned int (*Config::current_hash_function)(unsigned int) = kennys_hash_16;

回答by Fred Foo

No, these are called function pointers.

不，这些被称为函数指针。

unsigned int (*fp)(unsigned int) = func_1;

回答by mkaes

You could also use function either from the c++0x or from boost. That would be

您还可以使用来自 c++0x 或 boost 的函数。那将是

boost::function<int(int)>

and then use bind to bind your function to this type.

然后使用 bind 将您的函数绑定到此类型。

Have a look hereand here

看看这里和这里

Ok here would be a example. I hope that helps.

好的，这将是一个例子。我希望这有帮助。

int MyFunc1(int i)
{
    std::cout << "MyFunc1: " << i << std::endl;
    return i;
}

int MyFunc2(int i)
{
    std::cout << "MyFunc2: " << i << std::endl;
    return i;
}

int main(int /*argc*/, char** /*argv*/)
{
    typedef boost::function<int(int)> Function_t;

    Function_t myFunc1 = boost::bind(&MyFunc1, _1);
    Function_t myFunc2 = boost::bind(&MyFunc2, _1);

    myFunc1(5);
    myFunc2(6);
}

回答by 0x476f72616e

From C++11 you can use std::functionto store functions. To store function you use it as follsonig:

从 C++11 开始，您可以使用std::function来存储函数。要存储功能，您可以将其用作 follsonig：

std::function<return type(parameter type(s))>

std::function<返回类型（参数类型）>

as an example here it is:

作为一个例子，它是：

#include <functional>
#include <iostream>

int fact (int a) {
    return a > 1 ? fact (a - 1) * n : 1;
}

int pow (int b, int p) {
    return p > 1 ? pow (b, p - 1) * b : b;
}

int main (void) {
    std::function<int(int)> factorial = fact;
    std::function<int(int, int)> power = pow;

    // usage
    factorial (5);
    power (2, 5);
}

回答by Naszta

typedef unsigned int (*PGNSI)(unsigned int);

PGNSI variable1 = func_1;
PGNSI variable2 = func_2;

回答by Alexander Gessler

unsigned int (* myFuncPointer)(unsigned int) = &func_1;

However, the syntax for function pointers is awful, so it's common to typedefthem:

然而，函数指针的语法很糟糕，所以它们很常见typedef：

typedef unsigned int (* myFuncPointerType)(unsigned int);
myFuncPointerType fp = &func_1;

回答by yasouser

IF you have Boostinstalled, you can also check out Boost Function.

如果您安装了Boost，您还可以查看Boost Function。