SHIFTOUT MSB/LSBFIRST clarification

[iw2fvo]

good day to all in this forum,
I am using PicBasic pro to load an 8 bit shift register and I wrote a very simple program like this:

INCLUDE "MODEDEFS.BAS"
DT var byte
DT=128 ' decimal
SHIFTOUT H_DATA,H_CLK,msbfirst ,[DT\8]

Then I do suppose that the MSB ( in this case "1" ) is transmitted first and then "0000000" is transmitted after it.
By looking to the output stream with a scope, I find that the bit at logic level = high ( dec 128) is shown to the left side of the scope and not at the right side of the scope display.
I am just a little bit confuesd about that: any clarification please ?
Thanks in advance.
bye

[peterdeco1]

I haven't done this in a long time but here is a snippet of code I used to send by SPI to a device:

Low PORTA.3 : Pause 10 'CHIP SELECT
ShiftOut PORTC.5,PORTC.3,0,[9254\16] 'playconfig0
ShiftOut PORTC.5,PORTC.3,0,[114\8] '1st play byte
High PORTA.3 : Pause 10 'deselect - end of word

The 0 after portc.3 means send code LSB. A 1 would send it MSB.

[iw2fvo]

thanks a lot,
Is your SPI device set to receive MSBfirst or LSBfirst please ?
Bye

[richard]

time on a scope flows from left to right

[iw2fvo]

I am so sorry Richard:
if the time flows from left to right, then the first bit sent should be on the right and not to the left of the scope !
Where am I wrong please ?
Bye

[richard]

which second was recorded first in time sequence ?