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我试图从语法的角度理解 C 如何处理复杂类型定义的底层机制。

考虑以下示例（问题末尾包含参考资料）。

typedef int (*p1d)[10];

是正确的声明，即这里的 p1d 是一个指向 10 个整数的数组的指针，就像它在使用 Array 类型的声明下一样。请注意，这不同于

typedef int *p1d[10];

这将使 p1d 成为一个包含 10 个指向 int 类型的指针的数组的名称。

因此，如果我考虑两个示例的运算符优先级（我将重写它们）：

int *p1d[10]; // Becomes ...
int* p1d[10];

因此，从左到右阅读并使用运算符优先级，我得到：“指向 int 类型的指针，名为 p1d，大小为 10”，这是错误的。至于其他/第一种情况：

int (*p1d)[10];

我读为“p1d 是一个指针，类型为 int，并且是一个包含 10 个此类元素的数组”，这也是错误的。

有人可以解释用于确定这些 typedef 的规则吗？我也想将它们应用于函数指针，我希望这个讨论也能解释const强制转换背后的逻辑（即：指向常量数据的指针与指向变量数据的 const 指针）。

谢谢你。

参考：

  1. C 教程：指向数组的指针：http ://www.taranets.net/cgi/ts/1.37/ts.ws.pl?w=329;b=285
  2. 运算符优先级：http ://www.swansontec.com/sopc.html
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2 回答 2

14

我的一位教授写了这个小指南来阅读这些声明。给它一个阅读，它会值得你花时间，并希望回答任何问题。

所有功劳归功于 Rick Ord ( http://cseweb.ucsd.edu/~ricko/ )

The "right-left" rule is a completely regular rule for deciphering C
declarations.  It can also be useful in creating them.

First, symbols.  Read

     *      as "pointer to"         - always on the left side
     []     as "array of"           - always on the right side
     ()     as "function returning"     - always on the right side

as you encounter them in the declaration.

STEP 1
------
Find the identifier.  This is your starting point.  Then say to yourself,
"identifier is."  You've started your declaration.

STEP 2
------
Look at the symbols on the right of the identifier.  If, say, you find "()"
there, then you know that this is the declaration for a function.  So you
would then have "identifier is function returning".  Or if you found a 
"[]" there, you would say "identifier is array of".  Continue right until
you run out of symbols *OR* hit a *right* parenthesis ")".  (If you hit a 
left parenthesis, that's the beginning of a () symbol, even if there
is stuff in between the parentheses.  More on that below.)

STEP 3
------
Look at the symbols to the left of the identifier.  If it is not one of our
symbols above (say, something like "int"), just say it.  Otherwise, translate
it into English using that table above.  Keep going left until you run out of
symbols *OR* hit a *left* parenthesis "(".  

Now repeat steps 2 and 3 until you've formed your declaration.  Here are some
examples:

     int *p[];

1) Find identifier.          int *p[];
                                  ^
   "p is"

2) Move right until out of symbols or right parenthesis hit.
                             int *p[];
                                   ^^
   "p is array of"

3) Can't move right anymore (out of symbols), so move left and find:
                             int *p[];
                                 ^
   "p is array of pointer to"

4) Keep going left and find:
                             int *p[];
                             ^^^
   "p is array of pointer to int". 
   (or "p is an array where each element is of type pointer to int")

Another example:

   int *(*func())();

1) Find the identifier.      int *(*func())();
                                    ^^^^
   "func is"

2) Move right.               int *(*func())();
                                        ^^
   "func is function returning"

3) Can't move right anymore because of the right parenthesis, so move left.
                             int *(*func())();
                                   ^
   "func is function returning pointer to"

4) Can't move left anymore because of the left parenthesis, so keep going
   right.                    int *(*func())();
                                           ^^
   "func is function returning pointer to function returning"

5) Can't move right anymore because we're out of symbols, so go left.
                             int *(*func())();
                                 ^
   "func is function returning pointer to function returning pointer to"

6) And finally, keep going left, because there's nothing left on the right.
                             int *(*func())();
                             ^^^
   "func is function returning pointer to function returning pointer to int".


As you can see, this rule can be quite useful.  You can also use it to
sanity check yourself while you are creating declarations, and to give
you a hint about where to put the next symbol and whether parentheses
are required.

Some declarations look much more complicated than they are due to array
sizes and argument lists in prototype form.  If you see "[3]", that's
read as "array (size 3) of...".  If you see "(char *,int)" that's read
as "function expecting (char *,int) and returning...".  Here's a fun
one:

                 int (*(*fun_one)(char *,double))[9][20];

I won't go through each of the steps to decipher this one.

Ok.  It's:

     "fun_one is pointer to function expecting (char *,double) and 
      returning pointer to array (size 9) of array (size 20) of int."

As you can see, it's not as complicated if you get rid of the array sizes
and argument lists:

     int (*(*fun_one)())[][];

You can decipher it that way, and then put in the array sizes and argument
lists later.

Some final words:

It is quite possible to make illegal declarations using this rule,
so some knowledge of what's legal in C is necessary.  For instance,
if the above had been:

     int *((*fun_one)())[][];

it would have been "fun_one is pointer to function returning array of array of
                                          ^^^^^^^^^^^^^^^^^^^^^^^^
pointer to int".  Since a function cannot return an array, but only a 
pointer to an array, that declaration is illegal.


Illegal combinations include:

     []() - cannot have an array of functions
     ()() - cannot have a function that returns a function
     ()[] - cannot have a function that returns an array

In all the above cases, you would need a set of parens to bind a *
symbol on the left between these () and [] right-side symbols in order
for the declaration to be legal.

Here are some legal and illegal examples:

int i;                  an int
int *p;                 an int pointer (ptr to an int)
int a[];                an array of ints
int f();                a function returning an int
int **pp;               a pointer to an int pointer (ptr to a ptr to an int)
int (*pa)[];            a pointer to an array of ints
int (*pf)();            a pointer to a function returning an int
int *ap[];              an array of int pointers (array of ptrs to ints)
int aa[][];             an array of arrays of ints
int af[]();             an array of functions returning an int (ILLEGAL)
int *fp();              a function returning an int pointer
int fa()[];             a function returning an array of ints (ILLEGAL)
int ff()();             a function returning a function returning an int
                                (ILLEGAL)
int ***ppp;             a pointer to a pointer to an int pointer
int (**ppa)[];          a pointer to a pointer to an array of ints
int (**ppf)();          a pointer to a pointer to a function returning an int
int *(*pap)[];          a pointer to an array of int pointers
int (*paa)[][];         a pointer to an array of arrays of ints
int (*paf)[]();         a pointer to a an array of functions returning an int
                                (ILLEGAL)
int *(*pfp)();          a pointer to a function returning an int pointer
int (*pfa)()[];         a pointer to a function returning an array of ints
                                (ILLEGAL)
int (*pff)()();         a pointer to a function returning a function
                                returning an int (ILLEGAL)
int **app[];            an array of pointers to int pointers
int (*apa[])[];         an array of pointers to arrays of ints
int (*apf[])();         an array of pointers to functions returning an int
int *aap[][];           an array of arrays of int pointers
int aaa[][][];          an array of arrays of arrays of ints
int aaf[][]();          an array of arrays of functions returning an int
                                (ILLEGAL)
int *afp[]();           an array of functions returning int pointers (ILLEGAL)
int afa[]()[];          an array of functions returning an array of ints
                                (ILLEGAL)
int aff[]()();          an array of functions returning functions
                                returning an int (ILLEGAL)
int **fpp();            a function returning a pointer to an int pointer
int (*fpa())[];         a function returning a pointer to an array of ints
int (*fpf())();         a function returning a pointer to a function
                                returning an int
int *fap()[];           a function returning an array of int pointers (ILLEGAL)
int faa()[][];          a function returning an array of arrays of ints
                                (ILLEGAL)
int faf()[]();          a function returning an array of functions
                                returning an int (ILLEGAL)
int *ffp()();           a function returning a function
                                returning an int pointer (ILLEGAL)
于 2013-10-21T18:53:00.337 回答
3

稍微简化一下 KepaniHaole 的规则，可以归结为：

  1. 找到最左边的标识符
  2. 努力解决问题，记住缺少括号、函数调用()和[]绑定之前的显式分组*。
  3. 递归地应用于任何函数参数。

因此，T *a[]是指针 tp 的数组T，T (*a)[]是指向 的数组的指针T，T *f()是返回指向T,T (*f)()的指针的函数， 是指向返回 的函数的指针T。

举一个让很多人胃口大开的例子，我们可以看一下 POSIXsignal函数的原型：

void (*signal( int sig, void (*func)( int )))( int );

读作

       signal                                          -- signal
       signal(                             )           -- is a function with
       signal(     sig                     )           -- parameter sig
       signal( int sig                     )           --   of type int
       signal( int sig,        func        )           -- and parameter func
       signal( int sig,      (*func)       )           --   of type pointer to
       signal( int sig,      (*func)(     ))           --   a function with
       signal( int sig,      (*func)( int ))           --     an int parameter
       signal( int sig, void (*func)( int ))           --   returning void
     (*signal( int sig, void (*func)( int )))          -- returning a pointer to
     (*signal( int sig, void (*func)( int )))(     )   --   a function with
     (*signal( int sig, void (*func)( int )))( int )   --     an int parameter
void (*signal( int sig, void (*func)( int )))( int )   --   returning void
于 2013-10-21T20:54:38.047 回答