I am trying to wedge/hack some existing 6502 code that does a BMI instruction. Note that I don't want to insert any new code into the existing code - I'm replacing 3 bytes where possible with a JSR to my patch routine.

e.g.


I need to look ahead a bit further in my patch routine, but I need to restore Y, so I'm currently doing this:



I apologise if I'm tired and talking complete rubbish but I am correct in that the DEY instruction will clear/set the negative flag based on the result of Y, yes?

Assuming I'm right about this, then how can I re-test the value in A so that the N flag becomes set according to the value of A?

I wonder if ORA #$00 would do the trick?

cmp #0

will set the N flag according to the value in A.

That is all I can say without additional requirements.

Perfect! I see ora #$00 seems to work too

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What's an additional VIA among friends, anyhow?

That's a cool idea - which will be helpful in the future. Unfortunately there is a larger context here so I have to preserve the registers as best as I can.. I forgot to restore the contents of A before and got into trouble.. but PHA/PLA did the trick.

Be aware that PLA clobbers the flags...

I have to say, that seems like a poor data structure. The 2 is being used as a signal byte for the next byte and the 2 is basically discarded. What a waste of a byte. If it is not a 2, then use that value. If it is a 2, then check the next byte to see if it is negative. A lot of times they can be combined and just use the byte with the hi-bit value with what I call a signal bit. The hi-bit is the signal to do something else. Then it is just a matter of BMI or BPL.

(I'm thinking PHP, PLP is a natural way to save status. Of course it costs some cycles, but that will not usually matter.)

IIRC ...
CMP #0 -- will set/reset the S, Z and C flags

AND #$FF, ORA #0 -- will set/reset the S and Z flags, but not the C flag

For completeness, BIT #$80 -- will set/reset the Z flag based on the content of bit7, won't affect S or C (though not what you need here, since you need to restore the S flag to what it was when A was loaded)

No, BIT will set the N(S?) and V flags to the bit 7 and bit 6 values of the tested location, and set the Z flag to the result of ANDing the accumulator with the value from the memory location. C is not changed.

Edit: sorry I described BIT abs/zp. Bruce wrote about BIT immediate. He's very likely right. BIT immediate is a CMOS opcode if I remember correctly, so will work on 65c02 and up, but not on an NMOS 6502

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In the opcode summary I was looking out, it noted that BIT immediate is the only opcode where the immediate sets the flags differently from the other address modes. The operand of the BIT immediate would have been the source of the Sign and Overflow flags, and when they designed the 65C02 instructions, they must have decided that it was better to leave them alone than to set/clear them to a known state based on the immediate operand.

There is one use case they did not foresee or chose to ignore: self-modifying the immediate operand of a BIT instruction to set or clear the N and V flags.

Yeah, I reckon if someone wants to store a four way state, they can store it in the high two bits of a STATE value in zero page, and use "BIT STATE" to set the N and V flags simultaneously. So they may have figured that as already provided for by the NMOS 6502 instruction set.