铸造为一个函数指针？ [英] Cast to function pointer?

问题描述

所遇到如下所示的code线
我想这可能是一个投地返回void并接受一个空指针的函数指针，是正确的？

 （无效（*）（无效*））SGENT_1_calc
 

解决方案

是的，这是正确的。我觉得这不是很可读的，所以我建议在声明函数的签名必须指出：

 无效的typedef sigrout_t（无效*）;
 

^{我也有编码约定的以 rout_t 结束的类型是这种类型的函数签名。否则你可能名字，因为 _t 为后缀的由Posix的保留}

在我后面很铸造，也许叫它像

 （（sigrout_t *）SGENT_1_calc）（someptr）;
 

Have come across the line of code shown below I think it may be a cast to a function pointer that returns void and takes a void pointer, is that correct?

(void (*)(void *))SGENT_1_calc

解决方案

Yes it is correct. I find that not very readable, so I suggest declaring the signature of the function to be pointed:

 typedef void sigrout_t(void*);

^{I also have the coding convention that types ending with rout_t are such types for functions signatures. You might name it otherwise, since _t is a suffix reserved by Posix}

latter on I am casting, perhaps to call it like

 ((sigrout_t*) SGENT_1_calc) (someptr);

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