Release Rx, Tx resource back to Pin

[lgjonathan]

Hi,
I have a use case that I need to send a pulse with UART before communicating.
So my approach was to call set_high, set_low and delay to send a pulse.

But once the USART consumes the pin, I couldn't release the pin back to output pin.
From what I see, Serial::new() and Serial::release() is not bidirectional because Tx takes the ownership of the underlying pin.

I was wondering if there's a way to release Tx back to Pin?
(so I can call into_push_pull_output)

Thanks in advance

[burrbull]

You just need to unwrap Tx pin enum. https://docs.rs/stm32f4xx-hal/latest/stm32f4xx_hal/gpio/alt/usart1/enum.Tx.html#impl-TryFrom%3CTx%3COtype%3E%3E-for-Pin%3C'A',+9,+MODE%3E

let (usart, (txpin, rxpin)) = serial.release();
let txpin: gpio::PA9<Output> = txpin.try_into().unwrap(); // Now it is already in `Output<PushPull>>`

[lgjonathan]

Works! Much thanks