Why is LDAA 0,X Equivalent to LDAA 1,X+?

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LDAA 0,X loads the A accumulator with the value at the address in the X register plus an offset of 0. In contrast, LDAA 1,X+ loads the A accumulator with the value at the address in the X register plus an offset of 1, while also incrementing the X register after the operation. The confusion arises from the use of the offset in the second instruction, which is necessary for the intended operation. The first argument in the LDAA instruction specifies the offset to be added to the address in the X register. Understanding these differences clarifies why the two instructions, despite their similarities, are not interchangeable.
seang
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I cannot understand why the following bits are equivalent:

Code: 
LDAA 0,X
INCX
;and
LDAA 1,X+


The 1 is throwing me off. Since the X register is post incremented, why is it necessary? I would think that LDAA 0,X+ would be equivalent to the first bit of code. What am I missing?
 
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I don't program in HC 12 assembly, but is the first arguent to the LDAA instruction the register number?
 
Right, I was thinking that might be a problem.

LDAA 0,X loads the A accumulator with the address stored in the X register, plus the offset (0, in this case).

LDAA 1,X+ loads the A accumulator with the address in the X register, plus the offset (1 in this case), and then increments the X register.
 

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