SHIFTOUT MSB/LSBFIRST clarification

[iw2fvo]

thanks again,
I am sorry for not understand...
If I slow down the scope time base the dot goes from left to right.
So the first bit transmitted by the PIC is always to the right of the scope and the last transmitted by the PIC is shown to the left.
Am I right ?
Then the 9 seconds old bit is the first transmitted by the PIC.
The 0 secs old bit is the last transmitted by the PIC.
Is it right ?
Thanks

[iw2fvo]

sorry: I did not tell that I am talking about the PIC output .
Thanks

[richard]

completely wrong , the first bit is on the left

[iw2fvo]

could you please check in the real situation:

connect the PIC DATA output to scope ch_1
connect the PIC clock output to scope ch_2 and set the trigger to this channel.
Tranmit bin 00000011 ( dec 3 ) \8 selecting MSB FIRST. and insert a small delay in the main loop just for better scope triggering. My scope is a old style !

What do you see on the scope ? 00000011 or 11000000 ?
Or better: do you see "11" to the right side or to the left side of the scope ?

Thanks for helping
regards

[peterdeco1]

You will send 00000011.

[iw2fvo]

Hi Peter,
could you please clarify better ?
( my English is not a first language )
Thanks

[iw2fvo]

iT IS MY FAULT:
Richard explanation is correct as usual !
Thanks for all.
All the best
regards,
Bye

[Ioannis]

In the first post you said the number was 128 (10000000 in binary) so the MSB is indeed a '1'.

But on the last post you saw a number 3 (00000011 in binary) so the MSB is a '0'.

Ioannis