C 中单个结构成员的大小 [英] sizeof single struct member in C

问题描述

我试图声明一个依赖于另一个结构的结构.我想使用 sizeof 来保证安全/迂腐.

I am trying to declare a struct that is dependent upon another struct. I want to use sizeof to be safe/pedantic.

typedef struct _parent
{
  float calc ;
  char text[255] ;
  int used ;
} parent_t ;

现在我想声明一个与 parent_t.text 大小相同的结构 child_t.

Now I want to declare a struct child_t that has the same size as parent_t.text.

我该怎么做?(下面的伪代码.)

How can I do this? (Pseudo-code below.)

typedef struct _child
{
  char flag ;
  char text[sizeof(parent_t.text)] ;
  int used ;
} child_t ;

我用 parent_t 和 struct _parent 尝试了几种不同的方法，但我的编译器不接受.

I tried a few different ways with parent_t and struct _parent, but my compiler will not accept.

作为一个技巧，这似乎有效:

As a trick, this seems to work:

parent_t* dummy ;
typedef struct _child
{
  char flag ;
  char text[sizeof(dummy->text)] ;
  int used ;
} child_t ;

是否可以在不使用dummy的情况下声明child_t?

Is it possible to declare child_t without the use of dummy?

推荐答案

虽然使用 #define 定义缓冲区大小是一种惯用的方法，但另一种方法是使用这样的宏:

Although defining the buffer size with a #define is one idiomatic way to do it, another would be to use a macro like this:

#define member_size(type, member) sizeof(((type *)0)->member)

并像这样使用它:

typedef struct
{
    float calc;
    char text[255];
    int used;
} Parent;

typedef struct
{
    char flag;
    char text[member_size(Parent, text)];
    int used;
} Child;

我实际上有点惊讶 sizeof((type *)0)->member) 甚至允许作为常量表达式.很酷的东西.

I'm actually a bit surprised that sizeof((type *)0)->member) is even allowed as a constant expression. Cool stuff.

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