> 3; To illustrate how interesting that gets, consider Microsoft’s definition of the .NET modulus operator, which is: A Mod B is the equivalent of A – Int(A / B) * B + CLng(Math.Sign(A) Math.Sign(B)) * B, […] Efficient C Tips #7 â Fast loops Efficient C Tip #11 â Avoid passing parameters by using more small functions Efficient C Tip #13 â use the modulus (%) operator with caution […]. Minutes = seconds * MinutesMul << 32 and http://en.wikipedia.org/wiki/Double_dabble ) Here is my attempt for the 3-digits problem. All 3 CPUs divide by software, but ARM, being 32-bit, does it much faster, wrote division code myself in assembly, it is very easy, by the way. MOV R0,R1 This causes no loss of accuracy, because the number of seconds in a day, which we normally think of as 60*60*24 = 86400, can also be written as 128 * 675. Gauthier, 10, Jul 19. As a result it’s hardly surprising that code that uses the modulus operator can take a long time to execute. uint32_t days, hours, minutes, seconds; SecondsInDay = 24*60*60 Use the % Operator to Calculate Remainder in Division. ^ stime = time / 60UL; In computing, the modulo operation returns the remainder or signed remainder of a division, after one number is divided by another (called the modulus of the operation).. (here is the C code: `index = (index + 1) % 8;`, with index an uint8_t). I think this is flawless up to 102300 but above 999 the hundreds will not be a single digit, of course. Notwithstanding the ARM results, it’s clear that at least in this example, it’s possible to significantly speed up an algorithm by eliminating the modulus operator. days = (7 +97*a + ((93*a)>>10) â((59*a)>>17) ) >>16 I can’t see the advantage of doing this. on Tuesday, February 8th, 2011 at 9:21 am and is filed under Algorithms, Efficient C/C++, General C issues. I futher optimized by knowing the limited input bound and using remainders from a table from the computations. C = 2 in the first operation. If the binary value in any of the BCD columns is 5 or greater, add 3 to that value in that BCD column. MOV R1,R7. Another alternative to try is the div() function – this is a standard C function that calculates a division and modulus together. Wikipedia has a good table of the operators and their behavior. seconds = stime – (uint_fast16_t)minutes * 60U; Modifying a variable twice without sequence points in between is undefined behavior (ISO 9899:1999 6.5 Â§2). For example most (all?) uint8_t q2, q3; When this is done, it’s not usually long before one has to convert the ‘time’ into days, hours, minutes and seconds. Compared to your solution I dont’t need to divide by “big” numbers like 24UL*3600UL. b : c; parses as (std:: cout << a)? What kind of ‘running cycles’ statistical tools do you used? Thank you mr. Nigel. As a sidenote, any static code analyzer tool with MISRA-C will find both of these bugs implicitly, as the bugs are covered by the MISRA-C rules. This is what PC-Lint had to say: PC-lint for C/C++ (NT) Vers. Some compilers can do this for open-coded use of / and % too (gcc, at least recent-ish versions, does so). The ARM was nearly two orders of magnitude more cycle efficient than the MSP430 and AVR. Anything relying on undefined behavior is usually to be regarded as a severe bug, since the compiler is free to give -any- result. In C++, Modulus is performed using arithmetic operator %. We need to effectively divide by (24*60*60). “i & n – 1” is equivalent to “i % n” (Knuth 4A, 136). In theory, it should assist the compiler to perform the same optimisation that you manually do. I then subtracted one from the other to get the published results. If a function is available to give you both, this could speed you code The results however were a little surprising: Thus while this technique yielded a roughly order of two improvements for the AVR and MSP430 processors, it had essentially no impact on the ARM code.Â  Presumably this is because the ARM has native support for the modulus operation. The second operator yields 1. When this is done, the designer typically implements the time as a 32 bit variable containing the number of seconds since a particular date. _BLF ??divu32_a,??rA? As far as I know, there are instruction set extensions with division for some members of different MCU families. The modulus operator (also informally known as the remainder operator) is an operator that returns the remainder after doing an integer division.For example, 7 / 4 = 1 remainder 3. (note, this is pseudocode; 64-bit typecasts will be needed if written in C), days =( ( (seconds * DaysDecMul ) >> 32 ) + seconds * DaysIntMul) >> 32 The 7 will be lost by the large right shift unless the incoming number is very large., in which case it compensates for the imprecision of the integer approximation. lsd = rem[r]; q3 = (q >> 1) + (q >> 2); Here we can prevent this by first shifting away seven bits from the incoming 32-bit number of seconds. printf(“%d %d %d %d”,b,a++,a,++a); 2. shifts and ‘and’ are fast ways to div and % for powers of 2. did you know the intel pentium divide is not even used. In the case of a uint8_t, the value is stored in a 16 bits register anyway (16 bits being the native word size), and I believe that the promotion from uint8_t to uint16_t wouldn’t do anything. 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