Parameter Passing: Difference between revisions
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→Example 6502 Assembly
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Certain commands such as <code>INC</code>,<code>DEC</code>,<code>ASL</code>,<code>LSR</code>,<code>ROL</code>, and <code>ROR</code> inherently alter the memory address that is used as an argument. Thus their use on a memory address is pass-by-reference.
<syntaxhighlight lang="6502asm">
INC $20 ;add 1 to the value at memory address $20.
</syntaxhighlight>
However, opcodes that use a register as their primary operand are strictly pass-by-value.
<syntaxhighlight lang="6502asm">
;If the carry was set prior to this operation, also add an additional 1 to the accumulator.
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LDA $40 ;load the value stored at memory address $40 into the accumulator
ROR A ;rotate right the accumulator. This is only a copy. The actual value stored at $40 is unchanged.
</syntaxhighlight>
The main takeaway from this is that by default, the use of <code>LDA/LDX/LDY</code> commands to load from memory is pass-by-value. In order to actually update the memory location, you must store the value back into the source memory address.
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=====Example of out (result)=====
This routine reads the joystick output on Commodore 64 and stores it in the output variable <code>joystick1</code>.
<
ReadJoystick:
lda $DC00
ora #%11100000
sta joystick1
rts
</syntaxhighlight>
=====Example of in/out (mutable)=====
This one's tricky because of the way pointers work on the 6502. Unlike the other architectures of its day there are no "address registers" per se, but we can make our own using the zero page.
<
LDA #$03 ;load the low byte of the parameter's address
STA $20 ;store it in the low byte of our pointer variable.
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adc #$20 ; 0x20 = decimal 32
sta ($20),y ; overwrite the old value stored in $0003 with the new one.
rts
</syntaxhighlight>
===Example [[68000 Assembly]]===
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