Sure. libuv makes no claims about when exactly the data will be sent (if
ever), except that it will be sent at most once and will not be
re-attempted after the uv_write callback is fired. (the receiving side may
have even fewer guarantees on timing and duplication, depending on the
socket protocol connecting them)


On Tue, Jan 1, 2019 at 7:42 PM Iñaki Baz Castillo <[email protected]> wrote:

> Hi, trying to simplify, I do this:
>
> ------------------------------------------------------------
> // 1) Create a loop.
> uv_loop_t* loop = new uv_loop_t;
> uv_loop_init(loop);
>
> // 2) Create a pipe.
> uv_pipe_t* pipe = new uv_pipe_t;
> uv_pipe_init(::GetLoop(), (uv_handle_t*)pipe, 0);
>
> // 3) Send data over the pipe:
> uv_write(req, (uv_stream_t*)pipe, &buffer, 1);
>
> // 4) Run the loop:
> uv_run(loop, UV_RUN_DEFAULT);
> ------------------------------------------------------------
>
> And it works. This is, data is received in the other pipe side.
> However, it ++ALSO++ works even if I do not call uv_run(). Does it
> make sense?
>
>
> --
> Iñaki Baz Castillo
> <[email protected]>
>
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