檢查函數指針類型

Question

讓我們定義一個結構parser：

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

我有一個初始化函數：

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; 
} 

在策略函數定義的地方。好吧，我的興趣在於確定在編寫單元測試時使用了哪種策略。任何想法如何實現？

我在互聯網上看到is_pointer函數C++ 11，但我不認爲這會對我有幫助。

Answer 1

解析器是一個變量：

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

如果你想知道哪個策略是，你可以使用：

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

也許，你更喜歡存儲策略號：

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; 
    } 
} 

然後，你可以初始化解析器爲：

const int STRATEGY= parser_init(); 

Answer 2

您可以比較函數指針

if(p.buffer_push_strategy == buffer_push_strategy1) 

見https://ideone.com/QQzL1c