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ciccioc74
Joined: 23 Mar 2010
Posts: 41
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 Posted: Thu Feb 13, 2025 3:41 am |
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| Quote: | This is why I prefer to give UART streams names that correspond to the
physical UART they use, so you 'know' which port is involved. |
sorry could you give me an example? |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19765
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| Posted: Thu Feb 13, 2025 6:39 am |
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So the stream that connects to the RS485, and uses UART4, call something
line RS485_U4. Then the handler is for INT_RDA4. If you change to using
a different UART, change the stream name to correspond to what UART it
uses, and any discrepancy becomes obvious.
I suspect you are actually reading the wrong UART in one of your interrupt
handlers, and hence the lockup. |
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ciccioc74
Joined: 23 Mar 2010
Posts: 41
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| Posted: Thu Mar 13, 2025 11:53 am |
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I had set aside the bootloader issue to focus on other things, but now I'd like to understand where the problem is. Analyzing the .lst file, I noticed that something doesn't add up. Could you help me figure out what's wrong?
| Code: |
CCS PCD C Compiler, Version 5.119d, 1 13-mar-25 18:38
Compiler operating in Evaluation Mode
To obtain a fully enabled compiler visit www.ccsinfo.com/products |
in the ISR timer2:
| Code: | ................... #INT_TIMER2
.................... void my_timer2_isr (){
*
019FC: PUSH 42
019FE: PUSH 36
01A00: PUSH 32
01A02: MOV W0,[W15++]
01A04: MOV #2,W0
01A06: REPEAT #C
01A08: MOV [W0++],[W15++]
.................... if(++cont_1s>100){//0,5 seconds
01A0A: INC 0802
01A0C: MOV 802,W4
01A0E: MOV #64,W3
01A10: CP W3,W4
01A12: BRA GE,1A1A
.................... cont_1s=0;
01A14: CLR 802
.................... output_toggle(PIN_D15);
01A16: BCLR.B 2D9.7
01A18: BTG.B 2DD.7
.................... }
.................... }
01A1A: BCLR.B 84.7
01A1C: MOV #1A,W0
01A1E: REPEAT #C
01A20: MOV [--W15],[W0--]
01A22: MOV [--W15],W0
01A24: POP 32
01A26: POP 36
01A28: POP 42
01A2A: RETFIE |
focus on the line :01A1A: BCLR.B 84.7
in the ISR USART3:
| Code: | .................... #INT_RDA3
.................... void myRXisr(){
*
01A3A: PUSH 42
01A3C: PUSH 36
01A3E: PUSH 32
01A40: MOV W0,[W15++]
01A42: MOV #2,W0
01A44: REPEAT #C
01A46: MOV [W0++],[W15++]
.................... U3RXIF=0;
01A48: BCLR.B 8E.2
.................... do{
.................... temp= fgetc(UART3);
01A4A: CALL 1A2C
01A4E: MOV.B W0L,804
01A50: CLR.B 805
.................... output_toggle(LED3);
01A52: BCLR.B 2D1.5
01A54: BTG.B 2D5.5
.................... } while (kbhit(UART3));
01A56: BTSC.B 252.0
01A58: BRA 1A4A
.................... }
01A5A: BCLR.B 8E.2
01A5C: MOV #1A,W0
01A5E: REPEAT #C
01A60: MOV [--W15],[W0--]
01A62: MOV [--W15],W0
01A64: POP 32
01A66: POP 36
01A68: POP 42
01A6A: RETFIE |
focus on the 01A5A: BCLR.B 8E.2
but on the interrupt of serial4: | Code: |
.................... #INT_RDA4
.................... void myRXisr485(){
01A6C: PUSH 42
01A6E: PUSH 36
01A70: PUSH 32
01A72: MOV W0,[W15++]
01A74: MOV #2,W0
01A76: REPEAT #C
01A78: MOV [W0++],[W15++]
.................... U4RXIF=0;
01A7A: BCLR.B 8F.0
.................... // do{
.................... // temp= fgetc(UART4);
.................... // output_toggle(LED4);
.................... // } while (kbhit(UART4));
.................... }
....................
01A7C: BCLR.B 8F.0
01A7E: MOV #1A,W0
01A80: REPEAT #C
01A82: MOV [--W15],[W0--]
01A84: MOV [--W15],W0
01A86: POP 32
01A88: POP 36
01A8A: POP 42
01A8C: RETFIE |
the line 01A7C: BCLR.B 8F.0 clear 8F.0 bit but U4RXIF is located at 8E.8. could you explain to me why? |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19765
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| Posted: Fri Mar 14, 2025 2:24 am |
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16bit registers........
0x8e, if accessed as a 16bit register has bit 8 as U4RXIF. However you can
also perform an _8bit_ access to the next address up to access the same
bit.
BCLR.W 8E.8 == BCLR.B 8F.0 ***** not actually possible see below.
The data sheet describes the registers with 16bit values and addresses.
However the code to talk to the bit set/reset, uses 8bit addressing and
values.
The reason is that the BLCR.W version, can only be used on a W reg,
not on any memory address. So to do the same with word addressing,
you would have to read the 16bit word into a W reg, then use BCLR.W
on this register, and then write this back. Three instructions, where the bit
addresed 'byte' version does it in one. |
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ciccioc74
Joined: 23 Mar 2010
Posts: 41
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| Posted: Fri Mar 14, 2025 2:43 am |
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| Ok. Thank. |
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