如何解释以下typedef语句

Question

How can I interpret following typedef statement.

I already have some knowledge through type convertions. They are used to reinterpret the bytes behind a variable as another type. Example:

unsigned char* byte = (unsigned char*) (some signed char);

But now I see following statement:

typedef void (*uv_read_cb)(uv_stream_t* stream, ssize_t nread, uv_buf_t buf);

https://github.com/bodokaiser/libuv/blob/master/include/uv.h#L314

What totally confuses me are:

• there is no name of the typedef
• how can we do a type conversion on some arguments???

Has somebody an answer on these?

Answer 1

The typedef does have a name... uv_read_cb which is a pointer to a function that accepts a uv_stream_t* , a ssize_t and a uv_buf_t argument and returns void .

This allows you to add things like a pointer to a function in a structure, passing function pointers to functions, etc., while allowing you to define what type of function can be assigned to the variable or passed...

typedef void (*uv_read_cb)(uv_stream_t* stream, ssize_t nread, uv_buf_t buf);

struct myVTable
{
    uv_read_cb  uv_read_callback;
};

void myVFunction(  uv_stream_t* stream, uv_read_db callback )
{
     ssize_t length = 100;
     uv_buf_t buf;
     myVTable table;

     table.uv_read_callback = callback;

     table.uv_read_callback( stream, length, buf );

     // or you could alternatively use 'callback( stream, length, buf );'
}

Answer 2

It's a pointer to a function that gets uv_stream_t* , ssize_t and uv_buf_t , and returns void. uv_read_cb is also the type, and you can use it to define other functions with the same signature. For example:

uv_read_cb my_func;

Answer 3

此语句将uv_read_cb定义为指向返回void的函数的指针。