Running code in RAM

[Just Jeff]

What would happen if code was running in RAM and hit address $00ff? (assuming the stack was clear or saved)

[alex_79]

Unless that last instruction of the code jumps somewhere else, the cpu would continue reading the next instruction at the following address, that is $0100. But $0100 is a mirror of $00 in the VCS and that's in the area where TIA registers are mapped, so the program would interpret what it reads there (the collision registers) as code and would likely crash (or behave in an unintended way in any case)

Edited April 5, 2020 by alex_79

[NoLand]

Particularly, when reading from the TIA, there are

00  ...  CXM0P
01  ...  CXM1P
02  ...  CXP0FB 

Meaning, if there's no collision, the CPU will execute a BRK instruction ($00).

[JetSetIlly]

Quote

Particularly, when reading from the TIA, there are

00  ...  CXM0P
01  ...  CXM1P
02  ...  CXP0FB 

Meaning, if there's no collision, the CPU will execute a BRK instruction ($00).

 

It depends which RAM mirror you read from. In the case of the primary mirror, allowing the PC to get to $0100 will mean the next instruction loaded is an ORA command (byte code $01). That's without collisions. With collisions, the opcode will variously be $11, $21 or $31. None of which would be particularly useful.

 

As far as I can see, there's no useful mirror/opcode/collision combination that would result in a branch command or jump command. Which is a shame, because that might actually be useful.

 

 

Edited April 6, 2020 by JetSetIlly