Ich habe diesen Code:
1 | #include <avr/io.h> |
2 | #include <avr/eeprom.h> |
3 | #include <stdint.h> |
4 | |
5 | uint8_t EEMEM t[1024]; |
6 | int main(void) |
7 | {
|
8 | while(1) |
9 | {
|
10 | |
11 | }
|
12 | }
|
Die Build-Ausgabe ist:
1 | Program Memory Usage : 192 bytes 0,3 % Full |
2 | Data Memory Usage : 1024 bytes 25,0 % Full |
3 | EEPROM Memory Usage : 1024 bytes 50,0 % Full |
Warum landet der EEPROM-Teil im RAM?