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Jorge Nuno · Posted: Tue Feb 03, 2009 10:48 pm Post subject: 68k endianess question

When I want to program the word #$00E0 (appearing like that in a HEX editor) into a memory, at address 0, to be read by the 68k, where does the byte #$00 goes to? d15-d8 or d7-d0? Confused

Thanks

TmEE co.(TM) · Posted: Tue Feb 03, 2009 11:15 pm Post subject:

I just changed my flashing code when I found out that upper and lower bytes were swapped on my flashcart Razz

00 should be on 0 (D0 to D7), and E0 should be on 1 (D8 to D15)... I think, I may be wrong...

EDIT, no, the other way....
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Jorge Nuno · Posted: Tue Feb 03, 2009 11:24 pm Post subject:

TmEE co.(TM) · Posted: Tue Feb 03, 2009 11:33 pm Post subject:

I think so... 00E0 looks like 00E0 to 68K, while it would look like E000 to x86 or other little-endian things...

If the ROM won't work, you know you got things the wrong way Smile

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Jorge Nuno · Posted: Tue Feb 03, 2009 11:34 pm Post subject:

Chilly Willy · Very interested Joined: 17 Aug 2007 Posts: 1956

The 68000 is big endian. The value in a register 0x11223344 would be 0x11 0x22 0x33 0x44 in memory seen as bytes in a row from low address to high.

HardWareMan · Posted: Wed Feb 04, 2009 3:46 am Post subject:

Chilly Willy · Very interested Joined: 17 Aug 2007 Posts: 1956

Exactamundo! Big-endian data in a nutshell. The SH2 used by the 32X is also big-endian. Funny enough, the Z80 is not. 16 bit values used by the Z80 are little-endian. That's because the Z80 is based on Intel's 8080 line of CPUs, and all Intel CPUs are little-endian.