Unfamiliar C syntax?
c, libuv
Solution
It's just a type cast. `&server` gives the address of the `server` variable and `(uv_stream_t*)` casts the type of that address.
Based on the other information in your post, it seems that `server` is of type `uv_tcp_t`, but `uv_listen` wants a pointer to `uv_stream_t`. That's why you take the address of `server` and cast it to `uv_stream_t*`.
Note: this only makes sense because of how libuv defines `uv_tcp_t` and `uv_server_t` — in general you can't just cast pointer types to other pointer types and expect anything reasonable to happen.
Problem
I am trying to understand C code written by others and have come across a piece of code that I do not understand all of the syntax and the parts I don't understand I don't know how to search for. I would appreciate either an explanation or information on how to search this and other similar things I may encounter, or both. The code in question is the following line: ``` int r = uv_listen((uv_stream_t*) &server, 128, on_new_connection); ``` I understand that r is a variable that is being declared and initialized to the value returned by the function "uv_listen()", 128 is a literal int paramater, on_new_connection has to be a function pointer since that is the name of a call back function that is called. server is a variable of a custom type (uv_tcp_t) and with the & it is referring to the address of server. What I don't understand is the "(uv_stream_t*) &server". It looks like this is one of the parameters to the function. I could understand a function call that returned a value as a parameter but this doesn't look like a function call. "uv_stream_t" is another custom type that is defined in their code. I don't know if it is helpful in understanding what it means but the line of code is from sample code written to help in understanding how to use libuv.