如何使函数返回一个指向函数的指针?(c++)

I'm trying to make a function that takes a character, then returns a pointer to a function depending on what the character was. I just am not sure how to make a function return a pointer to a function.

我正在尝试创建一个函数，它取一个字符，然后返回一个指向一个函数的指针，这取决于字符是什么。我只是不知道如何让一个函数返回一个指向一个函数的指针。

13 个解决方案

#1

int f(char) {
    return 0;
}

int (*return_f())(char) {
    return f;
}

No, seriously, use a typedef :)

不，认真地说，使用typedef:)

#2

#include <iostream>
using namespace std;

int f1() {
    return 1;
}

int f2() {
    return 2;
}

typedef int (*fptr)();


fptr f( char c ) {
    if ( c == '1' ) {
        return f1;
    }
    else {
        return f2;
    }
}

int main() {
    char c = '1';
    fptr fp = f( c );
    cout << fp() << endl;
}

#3

Create a typedef for the function signature:

为函数签名创建一个类型定义:

typedef void (* FuncSig)(int param);

Then declare your function as returning FuncSig:

然后将函数声明为返回的函数:

FuncSig GetFunction();

#4

typedef void (*voidFn)();

void foo()
{
}

voidFn goo(char c)
{
    if (c == 'f') {
        return foo;
    }
    else {
        //..
    }
    // ..
}

#5

Here is how to do it without using a typedef:

下面是如何在不使用typedef的情况下做到这一点:

int c(){ return 0; }

int (* foo (void))(){  //compiles
return c;
}

#6

Check out this site - http://cdecl.org

看看这个网站——http://cdecl.org。

Helps you convert english to C declarations and back!

帮助您将英语转换为C声明和返回!

Cool Stuff!

酷炫的东西!

This link decodes the example in erikallen's answer. int (*return_f())(char)

这个链接解码了erikallen的答案中的例子。int(* return_f())(字符)

#7

Syntax for returning the function:

返回函数的语法:

return_type_of_returning_function (*function_name_which_returns_function)(actual_function_parameters) (returning_function_parameters)

Eg: Consider the function that need to be returned as follows,

考虑以下需要返回的函数，

void* (iNeedToBeReturend)(double iNeedToBeReturend_par)
{
}

Now the iNeedToBeReturend function can be returned as

现在，失去亲人的功能可以返回as

void* (*iAmGoingToReturn(int iAmGoingToReturn_par))(double)
{
   return iNeedToBeReturend;
}

I Felt very bad to learn this concept after 3 years of professional programming life.

经过3年的专业编程生涯，我对学习这个概念感到很难过。

Bonus for you waiting down for dereferencing function pointer.

您等待取消引用函数指针的好处。

C++ Interview questions

c++的面试问题

Example for function which returns the function pointer is dlopen in dynamic library in c++

函数返回函数指针的例子在c++的动态库中是dlopen

#8

Easiest way is to typedef the pointer-to-function type you want, and then use that

最简单的方法是定义您想要的pointerto -function类型，然后使用它。

typedef void (*fnptr_t)(int, int);
fptr_t myfunc(char *) { ....

#9

I prefer returning objects and call the operator(). This way your function can return an interface and all classes can inherit from this. That is, if you're using C++ and not C.

我更喜欢返回对象并调用操作符()。这样，函数就可以返回一个接口，所有的类都可以继承这个接口。也就是说，如果你用的是c++而不是C。

Then you can use the parametrized factor method to return the objects based on your input.

然后可以使用参数化因子方法根据输入返回对象。

#10

This is the code to show return of a function pointer. You need to define the "function signature" to return first:

这是显示函数指针返回的代码。您需要定义“函数签名”以首先返回:

int returnsOne() {
     return 1;
}

typedef int(*fp)();

fp returnsFPtoReturnsOne() {
    &returnsOne;
}

In your specific case:

在特定的情况下:

fp getFunctionFor(char code) {
    switch (code) {
        case 'a': return &functionA;
        case 'b': return &functionB;
    }
    return NULL;
}

#11

In C++11 you can use trailing return types to simplify the syntax, e.g. assuming a function:

在c++ 11中，你可以使用末尾返回类型来简化语法，例如假设有一个函数:

int c(int d) { return d * 2; }

This can be returned from a function (that takes a double to show that):

这可以从函数中返回(需要双精度来显示):

int (*foo(double e))(int)
{
    e;
    return c;
}

Using a trailing return type, this becomes a bit easier to read:

使用跟踪返回类型，这变得更容易阅读:

auto foo2(double e) -> int(*)(int)
{
    e;
    return c;
}

#12

Something like this

像这样的东西

#include <iostream>

typedef char (*fn_ptr_t)(char);

char a_fn(char c)
{
  return c + 1;
}

char b_fn(char c)
{
  return c + 2;
}

fn_ptr_t
return_function(char c)
{
  fn_ptr_t result = 0;

  switch (c)
 {
    case 'a':
      result = a_fn;
      break;
    case 'b':
      result = b_fn;
      break;
 }

 return result;
}

int
main()
{
  fn_ptr_t fn = return_function('a');

  std::cout << "a(l) = " << (fn)('l') << std::endl;

  return 0;
}

#13

I'm assuming C here (no objects) :) :

我假设这里是C(没有对象):

// Type of function which takes a char and returns an int:
typedef int (*Func)(char a);

// An example of the function you're trying to return and which does something
// with char:
int exampleFunc(char a)
{
    return (int)(a + 42);
}

// The function returning the pointer to a function:
Func *returnAfunc(void)
{
    return exampleFunc;
}