c - typedef 指针是如何工作的

Question

typedef int* ptr_t;
int target;
const ptr_t a = ⌖
*a = 6;            //OK
a = &target;       //<- error: assignment of read-only variable ‘a’

很明显，指针是常量，而不是指向的值。如果使用了#define，则相反。

将修饰符应用于在 typedef 中声明的指针的规则是什么？

举个实际例子，考虑代码 void (**foo)(void);

• 如何对类型进行类型定义，将顶级指针限定为 const(例如，指向硬件位置)，将下一个指针限定为 volatile(例如，可由独立硬件修改)指针功能？

• typedef void (**foo)(void) 如果这是我们必须使用的固定声明，如何在源代码中执行上述操作？

Answer 1

C 标准的第 6.7.5.1 节描述了差异。它给出了一个例子:

const int *ptr_to_constant;
int *const constant_ptr;

并说:

The contents of any object pointed to by ptr_to_constant shall not be modified through that pointer, but ptr_to_constant itself may be changed to point to another object. Similarly, the contents of the int pointed to by constant_ptr may be modified, but constant_ptr itself shall always point to the same location.

最后，该部分的第 4 点描述了 typedef 的使用。

The declaration of the constant pointer constant_ptr may be clarified by including a definition for the type ‘‘pointer to int’’.

  typedef int *int_ptr;
  const int_ptr constant_ptr;

declares constant_ptr as an object that has type ‘‘const-qualified pointer to int’’.