[Coco] Assembly language question: The stack

[William Astle]

The code in the interpreter is a little less straight forward than that, 
but the resulting behaviour is equivalent to a JSR to your program so 
that is a reasonable mental model to work with. I only mention this in 
case anyone starts looking at the disassembly of the EXEC command in the 
Unravelled series and is confused by it.

On 2018-10-25 10:55 AM, tim franklinlabs.com wrote:
>     The BASIC "EXEC" command has a command in it; JSR (address of your
>     program). After the JSR command, there's more code which is the
>     continuation of the BASIC interpreter. When you type EXEC $1234, the
>     CPU will get to the JSR mnemonic  and call your program. In doing so,
>     it puts the address of the next byte (plus the data bytes) after the
>     JSR command on the stack. This is the return address. Once pushed, the
>     Program counter gets loaded with the entry point of your binary code
>     and runs it. Your code should end with either an RTS or a PULS PC (same
>     as RTS). This will pull the return address off the stack and put it in
>     the program counter. This address is the point after the original JSR.
>     Basic then continues from that point.
> 
>     Simple, no?
> 
>       On October 25, 2018 at 11:03 AM Luis Fern?ndez
>       <[1]luis46coco at hotmail.com> wrote:
>       >So the mnemonic would be RTB?
>       No, RTS Normal
>       It depends on how the call to the Asembler started, normally the
>       stack retains the return address to the basic and when doing an RTS,
>       it returns to the basic after the call
>       --
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>       [2]Coco at maltedmedia.com
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> 
> References
> 
>     1. mailto:luis46coco at hotmail.com
>     2. mailto:Coco at maltedmedia.com
>     3. https://pairlist5.pair.net/mailman/listinfo/coco
>