Hi, I'm developing 28027 now but I soon found that this chip has only 2 h/w breakpoints, while one of these 2 is already reserved by systme ( True C$$EXIT (0x3F0FC5, exit.c, line 31) 0x0 (0x0) Terminate Program Execution Default Group Hardware ), so only left 1 for my application use, and this is tedious when debugging, esp when single step: you must delete the prevously set breakpoint to allow step! My question is: is there any way to release the breakpoint reserved by system and for my application use? pls advise, thanks in advance.
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