Why would it be necessary to perform two casts to

问题:

When looking at unix-socket, I came across this code:

let timeout = unsafe {
    let mut timeout: libc::timeval = mem::zeroed();
    let mut size = mem::size_of::<libc::timeval>() as libc::socklen_t;
    try!(cvt(libc::getsockopt(self.0,
                              libc::SOL_SOCKET,
                              kind,
                              &mut timeout as *mut _ as *mut _,
                              &mut size as *mut _ as *mut _)));
    timeout
};

I was curious about these lines in particular:

&mut timeout as *mut _ as *mut _,
&mut size as *mut _ as *mut _

Why is it necessary to perform two casts to a mutable raw pointer in a row? Why wouldn't it have been sufficient to only cast once?

回答1:

The timeout for example corresponds to a *mut c_void parameter:

pub unsafe extern fn getsockopt(sockfd: c_int, level: c_int, optname: c_int,
                                optval: *mut c_void, optlen: *mut socklen_t) -> c_int

The timeout in that file is defined as:

let mut timeout: libc::timeval = mem::zeroed();

So it's of type libc::timeval. Now let's consider:

&mut timeout as *mut _ as *mut _

First you have &mut timeout so that is of type &mut libc::timeval. Then you do as *mut _ to coerce it to a raw mutable pointer of an inferred type, which in this case is the same type of libc::timeval, so the full type so far is: *mut libc::timeval, which doesn't match the parameter type *mut c_void. The final as *mut _ again infers the target type, which is now the parameter type *mut c_void, so this finally coerces the *mut libc::timeval to a *mut c_void.