#206 How can I get the upper bytes of a 32-bit z80 address?

[Philipp Klaus Krause]

The .rel files generated by SDCC use 32-bit addresses for z80 (and related), 24 bits for stm8.

For stm8, #(_i >> 16) gives me the topmost byte of the 24-bit address. But for z80 I apparently get the least-significant byte of the address that way, i.e. as if the >> 16 wasn't even there. It looks to me like that when comparing the stm8 and z80 .rel files, though my understnading of hte asxxxx .rel format is limited; and I definitely see it in the final .ihx after linking.

[Janko Stamenović]

I don't know what's going on with Z80 "long" addresses. I've tried this code to get some idea:

long __at 0x12345678 hello;
unsigned f( void ) {
    return ( (unsigned long)&hello ) >> 16;
}

and I see the following generated for -mz80:

    .area _DATA
_hello  =   0x12345678
...
;hello.c:3: return ( (unsigned long)&hello ) >> 16;
    ld  de, #_hello + 2

which gives in the .lst:

                                     50 ;hello.c:3: return ( (unsigned long)&hello ) >> 16;
    00000000 11 7A 56         [10]   51     ld  de, #_hello + 2

567A value hex, which is 5678 + 2.

We should be careful when we change the behavior, however. AFAIK there is various existing SDCC code already somehow using banks on z80 platforms.

[Philipp Klaus Krause]

I'll try to check on sdcc-user if anyone relies on the current behaviour. But I also found a bug report about it: [bugs:#2930].

[Philipp Klaus Krause]

From the one reaction on sdcc-user, apparently no one relies on current >> 24 behavior. So I'd suggest to make it behave for ez80/rabbit/tlcs90 as it does for stm8; if we do it for all z80-related, that would fix [bugs:#2930].