宏:#define a(b)({... c;})意味着a(b)返回c吗? [英] macro: does #define a(b) ({... c;}) means a(b) returns c?

问题描述

我在Linux内核中有以下代码:

I have this code in linux kernel:

#define task_cred_xxx(task, xxx)                        
({                                                      
    __typeof__(((struct cred *)NULL)->xxx) ___val;  
    rcu_read_lock();                               
    ___val = __task_cred((task))->xxx;              
    rcu_read_unlock();                              
    ___val;                                         
})

我以前从未见过这样定义的宏，这是否意味着task_cred_xxx(task，xxx)返回___ val?

I never saw macro defined like this before, does that mean this is task_cred_xxx(task, xxx) returns ___val?

谢谢！

推荐答案

正确.它将返回___val.但是，像这样的块表达式是GNU扩展，实际上并不是C标准的一部分.

Correct. It will return ___val. However, block expressions like these are a GNU extension and isn't actually part of the C standard.

http://www.toofishes.net/blog/gcc-compound -statement-expressions/

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