没有成员编译错误 [英] has no member compilation error

问题描述

错误：'list_item_t'没有名为'state'的成员

任何有创意的想法如何让这段代码无需警告和错误而编译？

$ b $

  #if defined（_DEBUG_）
 #define ASSERT assert 
 #else / * _DEBUG_ * / 
＃定义ASSERT（exp）（（void）（exp））
＃endif` 
 
 typedef struct list_item {
 struct list_item * p_next; 
 struct list_item * p_prev; 
 #ifdef _DEBUG_ 
 int state; 
 #endif 
} list_item_t; 
 
 main（int argc，char * argv）
 {
 list_item_t p_list_item; 
 
 ASSERT（p_list_item.state == 0）; 
 
  

解决方案

Just #define ASSERT as

 ＃如果已定义（_DEBUG_）
 #define ASSERT断言
 #else 
 #define ASSERT（exp）（void）0 
 #endif 
  

请注意，这可能会改变其他代码点的行为，因为 ASSERT 不再评估其参数，但这就是人们期望它行为的方式。

或者执行 _DEBUG _ 构建，但这并不能解决问题，它只是避免它。

I have the following code and when I'm trying to compile it, I get an error:

error: ‘list_item_t’ has no member named ‘state’

Any creative ideas how to make this piece of code compile without warnings and erros?

 #if defined (_DEBUG_)
 #define ASSERT       assert
 #else                           /* _DEBUG_ */
 #define ASSERT( exp ) ((void)(exp))
 #endif`

typedef struct list_item {
        struct list_item *p_next;
        struct list_item *p_prev;
 #ifdef _DEBUG_
        int state;
 #endif
 } list_item_t;

main(int argc, char *argv)
{
    list_item_t p_list_item;

    ASSERT(p_list_item.state == 0);
}

解决方案

Just #define ASSERT as

 #if defined (_DEBUG_)
 #define ASSERT       assert
 #else                          
 #define ASSERT( exp ) (void)0
 #endif

Note that this may change the behaviour of other code spots because ASSERT no longer evaluates its argument, but that's how people expect it to behave anyway.

Or perform a _DEBUG_ build, but this doesn't resolve the problem, it just avoids it.

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