Dear Phillip,

Yes, you are right.

When the processor fetch the operator from memory in LCALL instruction he expects the 16 bits value be stored in a BIG ENDIAN like model.
The same is for the MOV DPTR,#data16
However the storage process is a compiler ( assembler ) task not a processor one.
I am tempted to say the 8051 is a half big-endian processor in this case.
The only situation when the processor manipulates by him self a two byte value is in the automatically push-pull of the PC in the stack ( memory) in an ISR or in a CALL-RET like instructions.
In this case the 8051 acts ( in my opinion ) as an LITTLE-ENDIAN processor.

I don’t know why Intel took this decision; only I know it is a cause of many mistakes generated by beginners and by old assembler programmers when they temped to manually put a value for PC in the stack ( or manually retrieve the PC value from stack)


Regards,
Cristian