How can I perform AND two times with 2 separate registers?

[infidelity]

In my routine, I am able to load a register, use AND, and the value is changed. But after that, ill do the same with another register, and the value does not change....

Example

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LDA $70	;46 is in $70
AND #$F0	;i use this for checking bits, I want to make the (6) in the value (46) to 0, so now the value in A is 40
STA $70		;40 into $70

Now here is where I get no change...

LDA $84	;95 is in $84
AND #$FD	;i use this cause I want the (5) in the value (95) to become (D) making 95 into 9D, but nothing happens, and 95 still remains...

[Shiru]

Your code gives correct result, as 0x95 'and' 0xfd is 0x95.

If you want to replace low nibble, you should 'and' 0xf0 then 'or' new value (like 0x05).

[tokumaru]

AND is used to clear bits, and OR (ORA in the 6502) is used to set bits. To set part of an arbitrary value (i.e. you don't know the value beforehand) you must first clear the bits you want to change with AND and then set them to the values you want with OR. The exceptions are when you want the modified bits to be all 0 (an AND alone can do it) or all 1 (an OR alone will do). If you want to flip bits (0 becomes 1 and 1 becomes 0) you can use a XOR (EOR on the 6502).

[infidelity]

@Shiru, your example worked thank you!

One more question How do you compare a bit? Id like to see if a hex value matches #8, and if it does, to branch to whatever i want.

[Shiru]

If you want compare a number, well,

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 ;value to test in A
 cmp #$08
 beq address ; go somewhere if A=$08

However, if you want to test a bit, say, it is bit 3, which is $08 bitmask, then just use AND:

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 ;value to test a bit in A
 and #$08
 bne address ;go somewhere if bit 3 is set, i.e. and result is non zero

[tokumaru]

Id like to see if a hex value matches #8, and if it does, to branch to whatever i want.

When you say an "hex value" do you mean a nibble? Like the "8" in $38 or $8D? If so, you need to clear the bits you don't care about and then compare normally:

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and #$0f ;keep only the lower nibble
cmp #$08 ;compare it to 8
beq LowerNibbleIs8 ;branch if equal

[infidelity]

I need to preserve the original left bits, while being able to properly compare then alter the right bits. I hope that makes sense.

[3gengames]

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LDA #$FF ;11111111 in binary.
AND #$F0 ;11110000 in binary. 11110000 is the result in this case.
LSR A
LSR A
LSR A
LSR A ;A is not the same as 0000**** where **** is the same as the top 4 bits of the result of AND #$F0 since we just moved it right 4 bits.
CMP CompareValue ;Do your compare.
;Branch

[tokumaru]

I need to preserve the original left bits, while being able to properly compare then alter the right bits. I hope that makes sense.

For that you'd typically have the value stored in RAM before clearing the irrelevant half, so that later you can load it back from RAM to compare the other half.

[infidelity]

Thank you everyone for your help.