uint16 累加_如何把一个uint16整数分解成两个字节并传输？
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Yeah, the second is possibly less portable , but can do the job efficiently in this case
2019-6-28 07:12:13
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Thakns!
i need clean asms,so i will take the second.
2019-6-28 07:28:20
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this declaration cost 0 instructions, possibly less portable ;
union timer {
uint16_t i2cTimer;
uint8_t ?val[2];
} ?i2ct ?;
2019-6-28 07:43:39
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yes,better idea to take advantage of union,clean in c ?too...
2019-6-28 07:58:38
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That's also true
2019-6-28 08:06:55
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Just curious. Could you post the disassembly for this line? I don't see how it can take 10 instructions, even for Free mode.
2019-6-28 08:14:18
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*(uint16_t*)val = i2cTimer; // assuming val is word-aligned edit: doesn't matter for XC8
If you're building a buffer, just copy i2cTimer diectly to where it should be in the buffer.
2019-6-28 08:34:14
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@NorthGuy:Thanks! all of your advice have been of great help!
2019-6-28 08:51:51
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Hi,1and0,below was asm lines i tried to restore,really confused,please kindly check if something was wrong:
5417 ;RSN-1601A_Functions.c: 126: ToSendDataBuffer[8] = i2cTimer;
5418 015B 0023 movlb 3 ; select bank3
5419 015C 085B movf _i2cTimer^(0+384),w
5420 015D 0022 movlb 2 ; select bank2
5421 015E 00A8 movwf 40
5422
5423 ;RSN-1601A_Functions.c: 127: ToSendDataBuffer[9] = i2cTimer>>8;
5424 015F 0023 movlb 3 ; select bank3
5425 0160 085C movf (_i2cTimer+1)^(0+384),w
5426 0161 0022 movlb 2 ; select bank2
5427 0162 00E3 movwf (??_CmdService^(0+256)+1)//_CmdService is the function name which contain the codes
5428 0163 0023 movlb 3 ; select bank3
5429 0164 085B movf _i2cTimer^(0+384),w
5430 0165 0022 movlb 2 ; select bank2
5431 0166 00E2 movwf ??_CmdService^(0+256)
5432 0167 0863 movf (??_CmdService^(0+256)+1),w
5433 0168 00E2 movwf ??_CmdService^(0+256)
5434 0169 01E3 clrf (??_CmdService^(0+256)+1)
5435 016A 0862 movf ??_CmdService^(0+256),w
5436 016B 00A9 movwf 41
5437
5438 ;RSN-1601A_Functions.c: 128: break;//i do above in a "case:",so "break" is here
5439 016C 0008 return
2019-6-28 09:10:59
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Use pointer conversion,only 4 instructions:
5417 ;RSN-1601A_Functions.c: 126: ToSendDataBuffer[8] = i2cTimer;
5418 015B 0023 movlb 3 ; select bank3
5419 015C 085B movf _i2cTimer^(0+384),w
5420 015D 0022 movlb 2 ; select bank2
5421 015E 00A8 movwf 40
5422
5423 ;RSN-1601A_Functions.c: 127: ToSendDataBuffer[9] = ((uint8_t *)&i2cTimer)[1];
5424 015F 0023 movlb 3 ; select bank3
5425 0160 085C movf (_i2cTimer^(0+384)+1),w
5426 0161 0022 movlb 2 ; select bank2
5427 0162 00A9 movwf 41
5428
5429 ;RSN-1601A_Functions.c: 128: break;
5430 0163 0008 return
2019-6-28 09:16:40
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Union Version:only?cost 2 instructions!!!(XC8,free mode,what happeded with you)
typedef union
{ uint16_t ?valUint16;
uint8_t ? ?byte[2];
} union_uint16_t;
extern union_uint16_t ?i2cTimer;
//
ASM lines:
5419 ;RSN-1601A_Functions.c: 126: ToSendDataBuffer[8] = i2cTimer.byte[0];
5420 015B 086E movf _i2cTimer^(0+256),w
5421 015C 00A8 movwf 40
5422
5423 ;RSN-1601A_Functions.c: 127: ToSendDataBuffer[9] = i2cTimer.byte[1];//cost 2 instructions!!!
5424 015D 086F movf (_i2cTimer^(0+256)+1),w
5425 015E 00A9 movwf 41
5426
5427 ;RSN-1601A_Functions.c: 128: break;
5428 015F 0008 return
2019-6-28 09:24:29
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The comparisons are not representative and depend mostly on the way the compiler allocates variables. If they're in different banks it takes 8 instructions to move 2 bytes. If they're in the same bank -?5 instructions (4 if the bank is already selected). Otherwise, it's all the same.
2019-6-28 09:41:57
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WTF...even Free mode cannot be that stupid!?Lines 5427 to 5435 are totally unnecessary. That disassembly does this:
uint16_t CmdService;
CmdService = i2cTimer;
CmdService >>= 8;
ToSendDataBuffer[9] = CmdService;
2019-6-28 09:56:44
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Hi 1and0,
i just built a simple prj of xc8 under free mode to test this,by simulator,the last line did not work at all,i am afraid it is a bug of XC8 ver1.38,you could also test it on you side.
#include
#include
void main(void) {
uint16_t u16t=0xABCD;
uint8_t val[2];
val[0]= u16t;//it works,val[0] is 0xCD
val[1]= (u16t>>8) ;//vars got lost......
return;
}
asm lines:
217 ;psect for function _main
218 07EF _main:
219
220 ;main.c: 12: uint16_t u16t=0xABCD;
221
222 ;incstack = 0
223 ; Regs used in _main: [wreg+status,2]
224 07EF 30CD movlw 205
225 07F0 00F3 movwf main@u16t
226 07F1 30AB movlw 171
227 07F2 00F4 movwf main@u16t+1
228
229 ;main.c: 13: uint8_t val[2];
230 ;main.c: 14: val[0]= u16t;
231 07F3 0873 movf main@u16t,w
232 07F4 00F0 movwf ??_main
233 07F5 0870 movf ??_main,w
234 07F6 00F1 movwf main@val
235
236 ;main.c: 15: val[1]= (u16t>>8) ;
237 07F7 0874 movf main@u16t+1,w
238 07F8 00F0 movwf ??_main
239 07F9 0870 movf ??_main,w
240 07FA 00F2 movwf main@val+1
241
242 ;main.c: 17: return;
2019-6-28 10:15:35
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Hi 1and0,is that possible to know result on you side? MPLAB x blue-screened my PC several times while compiling these days,maybe i need to re install them.
2019-6-28 10:27:02
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That is an invalid test.
If your variable is never used after being written (as you have done here),
and if you don't make your variable "volatile",
then the compiler is free to discard the writing code altogether as part of its optimisation.
2019-6-28 10:34:40
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in fact,if i just add a while(1); at the end,things are in order...
2019-6-28 10:50:52
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No, it looks like it is still there at line 240.
2019-6-28 11:07:31
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sorry,i just check the output result when debugging after adding the while(1);without review asm lines;
does that mean something wrong with XC8? or need update.
2019-6-28 11:16:55
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