为什么在这个接口中使用`typedef void * COMPLEX`? [英] Why is `typedef void * COMPLEX` used in this interface?
问题描述
我有一个程序，但我无法理解它是如何工作的.这是其中的一部分.我不明白行 typedef void *COMPLEX、命令 this 以及为什么使用 struct COMPLEX_IMPL.
I have a program and I can't understant how it works. Here is a part of it.
I don't understand the line typedef void *COMPLEX, the command this and why the struct COMPLEX_IMPL is being used.
#ifndef _COMPLEX_H
#define _COMPLEX_H
typedef void *COMPLEX;
COMPLEX NewCOMPLEX (double a, double b );
void DeleteCOMPLEX(COMPLEX this );
double GetA (COMPLEX this );
double GetB (COMPLEX this );
COMPLEX AddComplex (COMPLEX c1, COMPLEX c2, COMPLEX res);
COMPLEX MultComplex (COMPLEX c1, COMPLEX c2, COMPLEX res);
#endif /* _COMPLEX_H */
#ifndef _COMPLEX_H
#define _COMPLEX_H
typedef void *COMPLEX;
COMPLEX NewCOMPLEX (double a, double b );
void DeleteCOMPLEX(COMPLEX this );
double GetA (COMPLEX this );
double GetB (COMPLEX this );
COMPLEX AddComplex (COMPLEX c1, COMPLEX c2, COMPLEX res);
COMPLEX MultComplex (COMPLEX c1, COMPLEX c2, COMPLEX res);
#endif /* _COMPLEX_H */
<小时>
#include <stdio.h>
#include "complex.h"
struct COMPLEX_IMPL { double a; double b; };
double GetA(COMPLEX this) {
struct COMPLEX_IMPL *this_impl = (struct COMPLEX_IMPL*)this;
return this_impl->a;
}
推荐答案
typedef 定义类型的名称.所以
typedef defines a name for a type. So
typedef void *COMPLEX;
COMPLEX z;
相当于
void *z;
指针类型通常表示指针指向的数据类型.void * 是一个例外:它是一种拥有指针的方法，而无需说明它指向的值的类型.您可以自由地将任何类型的指针分配给 void * 指针并返回.
A pointer type normally indicates what kind of data the pointer points to. void * is an exception: it's a way to have a pointer without saying what the type of the value it points to is. You can freely assign any kind of pointer to a void * pointer and back.
void * 指针通常用于必须处理任何类型数据的通用库函数中.例如，考虑标准库函数memcpy:
void * pointers are normally used in generic library functions that must work with data of any type. For example, consider the standard library function memcpy:
void *memcpy(void *dest, const void *src, size_t n);
您向该函数传递一个指向任何类型对象的指针src，一个指向另一个对象(通常但不总是相同类型)的指针dest，以及要复制的字节数.该函数复制字节，它不关心字节的含义，因此传递两个指向未指定类型的指针就足够了.
You pass that function a pointer to an object of any type src, a pointer to another object (which is usually, but not always, of the same type) dest, and a number of bytes to copy. The function copies the bytes, it doesn't care what the bytes mean, so it's enough to pass two pointers-to-an-unspecified-type.
此处使用 void * 不是好的或常见的编程习惯.一个复数表示为它的实部和虚部:
The use of void * here is not good or common programming practice. A complex number is represented as its real part and its imaginary part:
struct COMPLEX_IMPL { double a; double b; };
典型的复数库会将其设为 COMPLEX 类型.
A typical complex number library would make this the COMPLEX type.
您发布的代码隐藏了 COMPLEX 类型的实现.复数被实现为包含两个 double 成员的结构这一事实仅在 complex.c 中显而易见.该库的用户只能看到 COMPLEX 是指向某物的指针.这是数据抽象的一种形式:隐藏数据的表示细节类型.但它做得不好:根据这个定义，任何指向任何东西的指针都可以分配给COMPLEX.通常的方法是使用一个不完整结构，它被声明并且明显是一个结构，但它的未指定成员.在 complex.h 中，你会写:
The code you posted hides the implementation of the COMPLEX type. The fact that complex numbers are implemented as a structure containing two double members is only apparent in complex.c. Users of the library only see that a COMPLEX is a pointer to something. This is a form of data abstraction: hiding the representation details of a data type. But it's poorly done: with this definition, any pointer to anything can be assigned to a COMPLEX. The normal way is to use an incomplete structure, which is declared and visibly a structure but whose members are not specified. In complex.h, you would write:
struct COMPLEX_IMPL;
typedef struct COMPLEX_IMPL *COMPLEX;
那样，合法创建COMPLEX_IMPL的唯一方法是通过complex.h提供的函数，但是COMPLEX类型的变量> 显然是一个指向 complex.c 中定义的复数表示的指针.
That way, the only way to legally create a COMPLEX_IMPL is through the functions provided by complex.h, but a variable of type COMPLEX is visibly a pointer to a representation of a complexe number as defined in complex.c.
哦，还有 this 是一个普通的变量名.
Oh, and this is an ordinary variable name.
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