C语言 转换为函数指针
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Cast to function pointer
提问by Toby
I have come across the line of code shown below.
我遇到了下面显示的代码行。
I think it may be a cast to a function pointer that returns void and takes a void pointer. Is that correct?
我认为它可能是对返回 void 并采用 void 指针的函数指针的强制转换。那是对的吗？
(void (*)(void *))SGENT_1_calc
回答by Basile Starynkevitch
Yes, it is correct. I find that not very readable, so I suggest declaring the signature of the function to be pointed:
是的，它是正确的。我发现这不是很可读，所以我建议声明要指向的函数的签名：
typedef void sigrout_t(void*);
I also have the coding conventionthat types ending with rout_tare such types for functions signatures. You might name it otherwise, since _tis a suffix reservedby POSIX.
我也有这样的编码约定，即以 结尾的rout_t类型是函数签名的类型。否则你可能名字，因为_t是一个后缀保留通过POSIX。
Later on I am casting, perhaps to call it like
后来我在铸造，也许称之为
((sigrout_t*) SGENT_1_calc) (someptr);
回答by bash.d
Yes, it is. The function should be looking like this
是的。该功能应该是这样的
void func(void*);
But the statement is missing a target, since a cast to nothing is useless. So it should be like
但是该语句缺少目标，因为强制转换为空是没有用的。所以它应该像
func = (void (*)(void *))SGENT_1_calc;
回答by Jeremy Cochoy
Yes, this is a function pointercast.
是的，这是一个函数指针转换。
Function pointer casts
函数指针强制转换
To help you with casting functions to pointers, you can define an alias for a function pointer type as follows:
为了帮助您将函数转换为指针，您可以为函数指针类型定义一个别名，如下所示：
typedef void void_to_void_fct(void*);
You can also define a type for a function that takes and returns values:
您还可以为接受和返回值的函数定义类型：
typedef int math_operator(int, int);
Later, you can store a function into a function pointer type like this:
稍后，您可以将函数存储到函数指针类型中，如下所示：
void mystery(void* arg) {
// do something nasty with the given argument
};
int add(int a, int b) {
return a + b;
}
void_to_void *ptr1 = mystery;
math_operator *ptr2 = add;
Sometimes, you have a function like print_str:
有时，您有这样的功能print_str：
void print_str(char* str) {
printf("%s", str);
}
and you want to store it in your function pointer that is agnostic to the argument type. You can then use a cast like this:
并且您想将其存储在与参数类型无关的函数指针中。然后，您可以使用这样的演员表：
(void (*)(void *))print_str
or
或者
(void_to_void_fct*)print_str
Why do we use function pointers?
为什么要使用函数指针？
Function pointers allow you to "store a function" inside a variable (indeed, you store the address of the function). This is very convenient when you want to allow some code to have diferent behavior depending on user input.
函数指针允许您在变量中“存储函数”（实际上，您存储的是函数的地址）。当您希望允许某些代码根据用户输入具有不同的行为时，这非常方便。
For exemple, suppose we have some data and some way to decode it. We could have the following structure to keep this information:
例如，假设我们有一些数据和一些解码方法。我们可以使用以下结构来保存这些信息：
typedef char* decoder_type(char*);
struct encoded_data {
char* data;
decoder_type *decoder_fct;
};
char* decoding_function_1(char* data) {
//...
char* decoding_function_2(char* data) {
//...
This allows storing both the data and the function to later use them together to decode the data.
这允许存储数据和函数以便以后一起使用它们来解码数据。
