检查函数指针类型

Question

让我们定义一个结构parser：

struct parser { 
    int (*buffer_push_strategy)(); 
    int (*escape_buffer_push_strategy)(); 
    int (*do_callback_strategy)(); 
}; 

我有一个初始化函数：

int parser_init() {  
    if (some_condition) { 
     parser->buffer_push_strategy = buffer_push_strategy1; 
     parser->escape_buffer_push_strategy = escape_buffer_push_strategy1; 
     parser->do_callback_strategy = do_callback_strategy1; 
    } 
    else { 
     parser->buffer_push_strategy = buffer_push_strategy2; 
     parser->escape_buffer_push_strategy = escape_buffer_push_strategy2; 
     parser->do_callback_strategy = do_callback_strategy2; 
    } 
return 0; 
} 

在策略函数定义的地方。好吧，我的兴趣在于确定在编写单元测试时使用了哪种策略。任何想法如何实现？

我在互联网上看到is_pointer函数C++ 11，但我不认为这会对我有帮助。

Answer 1

解析器是一个变量：

struct parserT { 
    int (*buffer_push_strategy)(); 
    int (*escape_buffer_push_strategy)(); 
    int (*do_callback_strategy)(); 
} parser; 

如果你想知道哪个策略是，你可以使用：

int strategy= (parser->buffer_push_strategy == buffer_push_strategy1) ? 1 : 2; 

也许，你更喜欢存储策略号：

int parser_init() {  
    if (some_condition) { 
     parser->buffer_push_strategy = buffer_push_strategy1; 
     parser->escape_buffer_push_strategy = escape_buffer_push_strategy1; 
     parser->do_callback_strategy = do_callback_strategy1; 
     return 1; 
    } 
    else { 
     parser->buffer_push_strategy = buffer_push_strategy2; 
     parser->escape_buffer_push_strategy = escape_buffer_push_strategy2; 
     parser->do_callback_strategy = do_callback_strategy2; 
     return 2; 
    } 
} 

然后，你可以初始化解析器为：

const int STRATEGY= parser_init(); 

Answer 2

您可以比较函数指针

if(p.buffer_push_strategy == buffer_push_strategy1) 

见https://ideone.com/QQzL1c