Initialize GPIO without change pin status. about onoff

[fivdi]

I wanna init a gpio instance(pin direction set to 'out') without change it's pin status('low' or 'high'),

If I understand the requirement correctly, I would expect something like the following to work:

const Gpio = require('onoff').Gpio;
const led = new Gpio(17, 'out', {reconfigureDirection: false});
...

[fivdi]

@elfive was is possible to make progress here? Did the proposed solution work?

[elfive]

I wanna init a gpio instance(pin direction set to 'out') without change it's pin status('low' or 'high'),

If I understand the requirement correctly, I would expect something like the following to work:

const Gpio = require('onoff').Gpio;
const led = new Gpio(17, 'out', {reconfigureDirection: false});
...

Sorry, the suggestion won’t work.
when direction set to "out", It will automatically set the pin status to LOW……

[fivdi]

when direction set to "out", It will automatically set the pin status to LOW……

This is not correct. When the reconfigureDirection option is set to false, the state of a pin that is already configured as an output (already exported) will not be modified.

What have you tried and what were the problems? Can you post a complete but short program that can be used to reproduce the problem please?

[elfive]

I’m writing a homebridge plugin,you can find my code here: https://github.com/elfive/homebridge-raspberry-simplegpio

[fivdi]

If you can post a complete but short program that can be used to reproduce the issue I can look into the problem. The short program shouldn't use homebridge. What I can't do is learn about homebridge to debug a problem that you are having.