SHIFTOUT MSB/LSBFIRST clarification

[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

[iw2fvo]

thanks for the response,
I had on my mund a shift register that will receive the serial stream from the PIC and I did associated the scope display to the shift register...
I was in error when I thought that the first bit on the left hand side of the scope is the last transmitted by the PIC: the first bit bit on the left of the scope is the first transmitted and the last bit to the right is the last transmitted.
Again, it was my fault .
Thanks for the assistance.
regards,Bye