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Question for the technically-minded


IRF

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Its the start of the higher memory area aka 32768 up (uncontended) but I don't really know the specifics other than lower memory timing is 'shared'. This is why things like special funky loaders have to live in higher (uncontended) memory so they can guarantee they will be 'active' when tape data is arriving.

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I might be muddling it up.

 

 

 

Contended Screen Ram

 

The block of RAM between &4000 and &7FFF is contented, that is access to the RAM is shared between the processor and the ULA. The ULA has priority access when the screen is being drawn.

 

Code that runs in or accesses contended memory will run slightly slower. For that reason it is recommended that speed critical routines (sprites, scrolling, etc) are not executed in this area of RAM.

 

http://www.animatez.co.uk/computers/zx-spectrum/memory-map/

 

http://scratchpad.wikia.com/wiki/Contended_memory

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It's clear that #8000 is the start of the uncontended (higher) memory.

 

But I'm still none the wiser as to why the program starts running at #8400?!

 

Is it hard-wired into the processor? i.e. do all Spectrum games have that starting point?

Edited by IRF
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But if I'm running game files on an emulator, with no loader, how does it know to start at #8400?

 

Are you asking how an emulator knows where to begin execution when loading a snapshot file?

 

If so, the answer is that the emulator takes the value of the program counter (PC) that is stored in the snapshot file. In a SNA file, PC is stored at the top of the stack (and must be popped off before execution begins). In a Z80 file, PC is stored in the header (like all the other register values). In an SZX snapshot, PC is stored in the ZXSTZ80REGS block (along with all the other register values).

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