writing 256 bytes to eeprom

[Archangel]

...then I got to wondering what they cost.
$40USD at Digikey! No thanks!
...

How can you put a price on <b>COOL</b> ? Especially for a one off Car project . . . Just imagine your GEO with all those cool . . .

[skimask]

How can you put a price on <b>COOL</b> ? Especially for a one off Car project . . . Just imagine your GEO with all those cool . . .

Right again...(not a GEO, just an old Chevy Sprint)

Cool Points > USD

[astouffer]

Yeah the $40 price probably keeps most hobbyists away from those displays. They really are about the size of a key on your keyboard.

This project is at the prototype stage now so changing pics is no problem. I'll order a few 18F2620 and start looking at the datasheet.

The '4620 has ~3900-ish bytes of ram. You should be able to double-buffer one screen in the ram easily, modify it there, then shift the whole thing out in one shot, if that's the way you want to go.

What command would write the screen into ram, writecode? The digits to be displayed are arrays of bytes? When an update is needed, the corresponding array of byte(s) is written to the specified ram location? I have an understanding of what needs done, now just putting it into code

[skimask]

Yeah the $40 price probably keeps most hobbyists away from those displays.

$40 doesn't keep this hobbyist away from anything...just don't have a use for them at the moment...Wait a minute! Since when did I have to have a use for something to buy something!!!

What command would write the screen into ram, writecode?

First thing you have to do is quit thinking about the on chip ram as some sort of special place to put data. Every variable you make is stored in on chip ram, every array, every byte, every x var byte : x = 10 x is ram, ram is x.
If you have an 18F2620 and you:
temp var byte[1023]
You have just allocated 1024 bytes of on chip to be declared as an array by the name of temp to be accessed with a statement such as temp[586] = 56.

[astouffer]

First thing you have to do is quit thinking about the on chip ram as some sort of special place to put data. Every variable you make is stored in on chip ram, every array, every byte, every x var byte : x = 10 x is ram, ram is x.
If you have an 18F2620 and you:
temp var byte[1023]
You have just allocated 1024 bytes of on chip to be declared as an array by the name of temp to be accessed with a statement such as temp[586] = 56.

Ah things are becoming less foggy now. As you might have guessed I'm new to programming in general. Thanks for being patient