凌乱的函数指针间pretation [英] Messy function pointer interpretation

问题描述

我碰巧遇到下面的函数指针。

I happen to come across the following function pointer.

char (*(*x())[])();

它看起来像以下格式的函数指针数组，但我看不出有什么的F - >（* X（））的意思。如何跨preT这混乱的函数指针？

It looks like an array of function pointer in the following format, but I can't see what f -> (*x()) means. How to interpret this messy function pointer?

char (*f[])();

ADDED

随着约翰·博德的帮助下，我做出了榜样如下：

ADDED

With John Bode's help, I make an example as follows.

#include <stdio.h>

char foo()    { return 'a'; }
char bar()    { return 'b'; }
char blurga() { return 'c'; }
char bletch() { return 'd'; }

char (*gfunclist[])() = {foo, bar, blurga, bletch};

char (*(*x())[])()
{
  static char (*funclist[4])() = {foo, bar, blurga, bletch};
  return &funclist;
}

int main() 
{
  printf("%c\n",gfunclist[0]());

  char (*(*fs)[4])();
  fs = x();
  printf("%c\n",(*fs)[1]()); 
}

我能得到预期的结果。

I could get the expected result.

smcho@prosseek temp2> ./a.out 
a
b

和，你可以找到一个更好的实施<一个href=\"http://stackoverflow.com/questions/4107934/messy-function-pointer-how-to-remove-the-warning\">here.

And, you can find a better implementation here.

推荐答案

我的一般程序是出来找在声明中最左边的标识符，然后工作，我的方式，记住， [] 和（）绑定之前 * （即 * F（）通常被解析为 *（F（））和 *一个[] 正常解析为 *（A []​​））。

My general procedure is to find the leftmost identifier in the declaration, and then work my way out, remembering that [] and () bind before * (i.e., *f() is normally parsed as *(f()) and *a[] is normally parsed as *(a[])).

因此​​，

          x           -- x
          x()         -- is a function
         *x()         -- returning a pointer
        (*x())[]      -- to an array
       *(*x())[]      -- of pointers
      (*(*x())[])()   -- to functions
 char (*(*x())[])();  -- returning char

将这样的野兽会是什么样的做法？

What would such a beast look like in practice?

char foo()    { return 'a'; }
char bar()    { return 'b'; }
char blurga() { return 'c'; }
char bletch() { return 'd'; }

/**  
 *           funclist           -- funclist
 *           funclist[]         -- is an array
 *          *funclist[]         -- of pointers
 *         (*funclist[])()      -- to functions
 *    char (*funclist[])()      -- returning char
 */    
char (*funclist[])() = {foo, bar, blurga, bletch};

这位前pression ＆放大器; funclist 将返回一个指针数组，所以

The expression &funclist will return a pointer to the array, so

char (*(*x())[])()
{
  return &funclist;
}

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