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mjdale2003
Joined: 10 Oct 2004
Posts: 21
Location: UK
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| With #OPT compress on 18f27j13 int16 shift problem |
 Posted: Mon Jan 25, 2016 5:54 am |
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Hi, I seem to be having a problem with shifting a 16bit integer when using the #opt compress switch
I am trying to store a 16bit integer in to internal flash.
i have tried both |= (1<<val) method and bit_set()
compiler version
PCH 5.051
the output to the function is
here is the function
| Code: |
#OPT compress
void save_16_eeprom(unsigned int16 val){
unsigned int16 dat_val;
value = read_ee16(active_output_cam);
01198: MOVFF AA8,C68
0119C: RCALL 0E68
0119E: MOVFF 02,C67
011A2: MOVFF 01,C66
// use bit_set as optimization
// changes this to a char shift
// for some reason even though
// printf shows val as 8 - 16 ???
//dat_val |= ((unsigned int16)1<<val);
bit_set(dat_val, val);
011A6: CLRF 03
011A8: MOVLW 01
011AA: MOVWF 02
011AC: MOVLB C
011AE: MOVF x64,W
011B0: MOVWF 00
011B2: BZ 11BE
011B4: BCF FD8.0
011B6: RLCF 02,F
011B8: RLCF 03,F
011BA: DECFSZ 00,F
011BC: BRA 11B4
011BE: MOVF 02,W
011C0: IORWF x66,F
011C2: MOVF 03,W
011C4: IORWF x67,F
write_ee16(active_output_cam, dat_val);
011C6: MOVFF AA8,C84
011CA: MOVFF C67,C86
011CE: MOVFF C66,C85
011D2: MOVLB 0
011D4: RCALL 0ECC
printf(TxData2, "save %lu %04X \n\r", val, dat_val); TXIE = 1;
011D6: MOVLW AE
011D8: MOVWF FF6
011DA: MOVLW BE
011DC: MOVWF FF7
011DE: MOVLW 00
011E0: MOVWF FF8
011E2: MOVLW 05
011E4: RCALL 10C8
011E6: MOVLW 10
011E8: MOVWF FE9
011EA: MOVFF C65,C69
011EE: MOVFF C64,C68
011F2: RCALL 10D2
011F4: MOVLW 20
011F6: MOVLB C
011F8: RCALL 122C
011FA: MOVLW 02
011FC: MOVLB C
011FE: MOVWF x68
01200: MOVLW 30
01202: RCALL 122C
01204: MOVLB C
01206: DECFSZ x68,F
01208: BRA 1200
0120A: MOVFF C66,C68
0120E: MOVLW 37
01210: MOVWF x69
01212: MOVLB 0
01214: RCALL 0F46
01216: MOVLW BB
01218: MOVWF FF6
0121A: MOVLW BE
0121C: MOVWF FF7
0121E: MOVLW 00
01220: MOVWF FF8
01222: MOVLW 03
01224: RCALL 10C8
01226: BSF F9D.4
*
0122C: MOVWF x73
0122E: MOVLB 0
01230: RCALL 0C66
01232: RETURN 0
update_eeprom();
01228: RCALL 1010
0122A: RETURN 0
}
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temtronic
Joined: 01 Jul 2010
Posts: 9162
Location: Greensville,Ontario
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| Posted: Mon Jan 25, 2016 7:30 am |
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I don't use that PIC but you should post a small,compilable program that shows us the problem. Most C guys don't know ASM. Could be a 'casting' problem or something 'before' your function.
Also post test data before and after as well as expected results
Jay |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19329
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| Posted: Mon Jan 25, 2016 8:47 am |
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Bit set has to do a rotation. The processor doesn't have a variable based bit_set function, so while a bit set on a fixed bit is efficient, a bit_set using a variable has to be done by rotating a mask.
I can't actually work out what you are trying to do?. If you are rotating a value to get bytes from it, then use the make8 function. Far more efficient than any rotation based method, since the compiler can just do a byte fetch. |
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mjdale2003
Joined: 10 Oct 2004
Posts: 21
Location: UK
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| Posted: Mon Jan 25, 2016 9:45 am |
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hi and thanks for the fast response, i have stripped down the function to a simple routine that assigns a value to the top byte prints, sets a bit then prints again, there is no reason this should fail surely
| Code: |
void save_pre_eeprom(unsigned int16 pre){
unsigned int16 value;
if(pre <=16){
value = 0xAA00;
printf(TxData2, "save %u %lu %04X \n\r", active_output, pre, value);
bit_set(value, 8);
printf(TxData2, "save %u %lu %04X \n\r", active_output, pre, value);
}
}
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and the output remains as
save 0 2 0000
save 0 2 0000
call it from main the "pre" value changes if i use another value but the "value" value is always 0000 even if i change it to a int32
the .lst file seems to set the bit but it is never displayed, could it be the printf function becoming corrupt after the #compress part of the compilation.
| Code: |
.................... void save_pre_eeprom(unsigned int16 pre){
.................... unsigned int16 value;
....................
.................... if(pre <=16){
*
01316: MOVLB C
01318: MOVF xE5,F
0131A: BTFSS FD8.2
0131C: BRA 137A
0131E: MOVF xE4,W
01320: SUBLW 10
01322: BTFSS FD8.0
01324: BRA 137A
....................
.................... value = 0xAA00;
01326: MOVLW AA
01328: MOVWF xE7
0132A: CLRF xE6
....................
.................... printf(TxData2, "save %u %lu %04X \n\r", active_output_cam, pre, value);
0132C: MOVLW DA
0132E: MOVWF FF6
01330: MOVLW C0
01332: MOVWF FF7
01334: MOVLW 00
01336: RCALL 137E
01338: MOVLW 30
0133A: RCALL 1246
0133C: MOVLB C
0133E: DECFSZ xE8,F
01340: BRA 1338
01342: RCALL 13AC
01344: MOVLW EA
01346: MOVWF FF6
01348: MOVLW C0
0134A: MOVWF FF7
0134C: MOVLW 00
0134E: RCALL 13B8
*
0137E: MOVWF FF8
01380: MOVLW 05
01382: RCALL 1024
01384: MOVFF B28,D09
01388: MOVLW 1B
0138A: RCALL 102E
0138C: RCALL 0EC0
0138E: MOVLW 20
01390: RCALL 1246
01392: MOVLW 10
01394: MOVWF FE9
01396: MOVFF CE5,CE9
0139A: MOVFF CE4,CE8
0139E: RCALL 1250
013A0: MOVLW 20
013A2: RCALL 1246
013A4: MOVLW 02
013A6: MOVLB C
013A8: MOVWF xE8
013AA: RETURN 0
013AC: MOVFF CE6,D09
013B0: MOVLW 37
013B2: RCALL 102E
013B4: RCALL 0F34
013B6: RETURN 0
013B8: MOVWF FF8
013BA: MOVLW 03
013BC: RCALL 1024
013BE: RETURN 0
....................
.................... bit_set(value, 8);
01350: MOVLB C
01352: BSF xE7.0
....................
.................... printf(TxData2, "save %u %lu %04X \n\r", active_output_cam, pre, value);
01354: MOVLW EE
01356: MOVWF FF6
01358: MOVLW C0
0135A: MOVWF FF7
0135C: MOVLW 00
0135E: RCALL 137E
01360: MOVLW 30
01362: RCALL 1246
01364: MOVLB C
01366: DECFSZ xE8,F
01368: BRA 1360
0136A: RCALL 13AC
0136C: MOVLW FE
0136E: MOVWF FF6
01370: MOVLW C0
01372: MOVWF FF7
01374: MOVLW 00
01376: RCALL 13B8
01378: MOVLB C
....................
.................... }
0137A: MOVLB 0
0137C: RETURN 0
.................... }
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RF_Developer
Joined: 07 Feb 2011
Posts: 839
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| Posted: Mon Jan 25, 2016 10:45 am |
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Its nothing to do with shifting or optimisation. Check your format specifiers:
From the manual:
| Quote: |
x Hex int (lower case)
X Hex int (upper case)
Lx Hex long int (lower case)
LX Hex long int (upper case)
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To format your int16, you need %LX. What your format is doing is formatting the lower byte and padding it to four characters with zeros: the upper byte is always going to be 00. |
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mjdale2003
Joined: 10 Oct 2004
Posts: 21
Location: UK
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| Posted: Mon Jan 25, 2016 11:27 am |
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| Well spotted thank you, silly me :/ |
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