??? 04/14/04 07:40 Read: times |
#68451 - RE: Hex file - question Responding to: ???'s previous message |
The short answer is "context". In this case (:05200000524541445966) the program loader is going to put a 52h in memory location 2000h. It frankly doesn't care if it's ASCII for an 'R', or if it's an opcode, or whatever. It just "knows" that it is going to write 01010010 in the 8 bit locations beginning at 2000h. It will then put 45h in 2001h, 41h in 2002h, 44h in 2003h and 59h in 2004h.
When the µ-controller executess the code, it will treat the 52 as whatever context requires. In other words, execution will begin at 0000h. There the µ-controller will encounter what it will assume is an op-code. It will then assume that the subsequent data bytes are the ones necessary for that opcode. If the opcode requires 3 bytes of data, it will assume that the next 3 bytes are those data and use them accordingly. It will then assume that the 4th byte is the next opcode. |
Topic | Author | Date |
Hex file - question | 01/01/70 00:00 | |
RE: Hex file - question | 01/01/70 00:00 | |
RE: Hex file - question![]() | 01/01/70 00:00 |