case 1:
static uint8_t i;
i % 3; // 3 is a signed number
case 2:
static uint8_t i;
i % 3u; // 3 is a unsigned number
With Microchip XC8 compiler, I found that case 1 needs more instruction(the program memory used is little increased). Why?
Note: 8-bit CPU. The XC8 compiler conforms to C89 standard.
assembly:
225: if(KEYbits.I2C == ON && h%5u == 0){
02F2 1ED2 BTFSS KEYbits, 0x5
02F3 2AF5 GOTO 0x2F5
02F4 2AF6 GOTO 0x2F6
02F5 2C75 GOTO 0x475
02F6 3005 MOVLW 0x5
02F7 00F0 MOVWF __pcstackCOMMON
02F8 3000 MOVLW 0x0
02F9 00F1 MOVWF hold
02FA 0850 MOVF h, W
02FB 00DE MOVWF 0x5E
02FC 01DF CLRF 0x5F
02FD 085E MOVF 0x5E, W
02FE 00F2 MOVWF product
02FF 085F MOVF 0x5F, W
0300 00F3 MOVWF multiplier
0301 318A MOVLP 0xA //here
0302 22B4 CALL 0x2B4 //here
0303 3180 MOVLP 0x0
0304 0870 MOVF __pcstackCOMMON, W
0305 0471 IORWF hold, W
0306 1D03 BTFSS STATUS, 0x2
0307 2B09 GOTO 0x309
0308 2B0A GOTO 0x30A
0309 2C75 GOTO 0x475
226: h = 0;
02B4 0834 MOVF TMR4_counter, W
02B5 07DE ADDWF 0x5E, F
02B6 0835 MOVF 0x35, W
02B7 3DDF ADDWFC 0x5F, F
02B8 0836 MOVF 0x36, W
02B9 3DE0 ADDWFC 0x60, F
02BA 0837 MOVF 0x37, W
02BB 3DE1 ADDWFC 0x61, F
02BC 0861 MOVF 0x61, W
02BD 00B7 MOVWF 0x37
02BE 0860 MOVF 0x60, W
02BF 00B6 MOVWF 0x36
02C0 085F MOVF 0x5F, W
02C1 00B5 MOVWF 0x35
02C2 085E MOVF 0x5E, W
02C3 00B4 MOVWF TMR4_counter
225: if(KEYbits.I2C == ON && h%5 == 0){
02F2 1ED2 BTFSS KEYbits, 0x5
02F3 2AF5 GOTO 0x2F5
02F4 2AF6 GOTO 0x2F6
02F5 2C75 GOTO 0x475
02F6 3005 MOVLW 0x5
02F7 00F0 MOVWF __pcstackCOMMON
02F8 3000 MOVLW 0x0
02F9 00F1 MOVWF hold
02FA 0850 MOVF h, W
02FB 00DE MOVWF 0x5E
02FC 01DF CLRF 0x5F
02FD 085E MOVF 0x5E, W
02FE 00F2 MOVWF product
02FF 085F MOVF 0x5F, W
0300 00F3 MOVWF multiplier
0301 3186 MOVLP 0x6 //here
0302 2663 CALL 0x663 //here
0303 3180 MOVLP 0x0
0304 0870 MOVF __pcstackCOMMON, W
0305 0471 IORWF hold, W
0306 1D03 BTFSS STATUS, 0x2
0307 2B09 GOTO 0x309
0308 2B0A GOTO 0x30A
0309 2C75 GOTO 0x475
226: h = 0;
0663 01F6 CLRF sign
14: if(dividend < 0) {
0664 1FF3 BTFSS multiplier, 0x7
0665 2E67 GOTO 0x667
0666 2E68 GOTO 0x668
0667 2E70 GOTO 0x670
15: dividend = -dividend;
0668 09F2 COMF product, F
0669 09F3 COMF multiplier, F
066A 0AF2 INCF product, F
066B 1903 BTFSC STATUS, 0x2
066C 0AF3 INCF multiplier, F
16: sign = 1;
066D 01F6 CLRF sign
066E 0AF6 INCF sign, F
066F 2E70 GOTO 0x670
The usual arithmetic conversions are performed on the operands of %. I am guessing that i is a signed type; the result of i % 3 may differ from the result of i % 3u.
For example, -1 % 3u is usually 0 because after conversion to unsigned int, -1 typically becomes one less than an even power of two, e.g. 65535 or 4294967295, which is always a multiple of 3.
But -1 % 3 is either -1 or 2 in C89 (it's implementation-defined which).
It stands to reason that the compiler might have to issue different assembly instructions for these two different cases.
The edit to the question doesn't change this; the uint8_t is promoted to signed int, and the optimizer might not realize that it could use the "better" assembly for both cases.