| ??? 02/27/07 20:47 Read: times Msg Score: -2 -2 Answer is Wrong |
#133883 - There is limit Responding to: ???'s previous message |
As said by Eric in his first reply, That you won't be able to get full 256 bytes of Ram but slightly lesser than that. Adding more to this, even if you use most of your ram, your program will behave abnormal from the required behavior.
Reason: Overwriting of certain variables! So its not only important to keep the space for stack but it is also necessary to keep room for variables, some gap is required. Saving ram space for constant variables can be done by declaring them in code space with compiler directive keyword [B]code[/B] e.g. [b]char code[/b] var1=0x50; So this save the use of ram, but still, it save 3 byte pointer pointing to this variable... So as i said.. ROOM is required. Hope this help!! Regards, Ajay |
| Topic | Author | Date |
| using 256 bytes of ram 0f 8052 in keil | 01/01/70 00:00 | |
| you can't, the most you can get is 248 (256 - one | 01/01/70 00:00 | |
| Keil internal ram | 01/01/70 00:00 | |
| resons for extending memory | 01/01/70 00:00 | |
| it is, and thus it takes many reads | 01/01/70 00:00 | |
| lookup | 01/01/70 00:00 | |
| not necessarily correct | 01/01/70 00:00 | |
| There is limit | 01/01/70 00:00 | |
| Try it if you think its wrong :) | 01/01/70 00:00 | |
| wht a bunch of gobbelygook! | 01/01/70 00:00 | |
| Qualifers | 01/01/70 00:00 | |
| Sloppy | 01/01/70 00:00 | |
| Maybe | 01/01/70 00:00 | |
| problem solved | 01/01/70 00:00 | |
| You Must Read | 01/01/70 00:00 | |
RTFMs | 01/01/70 00:00 | |
| What??? | 01/01/70 00:00 | |
| The solution could be as simple as: | 01/01/70 00:00 | |
| just note about size | 01/01/70 00:00 |