uint64_t p = (uint64_t) x * 2290649225u; // multiply x by ceiling(2^32 * 32 / 60) = 2290649225 = 0x88888889 Here’s one more idea for converting a 32-bit integer of seconds into days, hours, minutes and seconds. Modulus operator divide first operand from second and returns remainder. Thus an output of “My brain hurts” would in fact be ‘correct’. Line 5: Warning[Pa079]: undefined behavior: variable “a” (or a value reached by some form of indirection through it) is modified more than once without an intervening sequence point in this statement Prepare to multiply by 2^32 / (24 * 60*60) which we can pre-calculate. msb +=1; a = 10 will assign 10 in a: Relational operators If the first number (a) is non-zero and the second number is 0, it gives an error at compile time. Full marks for paying close attention Ralf. http://cc.davelozinski.com/c-sharp/use-the-modulus-operator-or-alternative. uint32_t r = x – (q * 60u); In this problem, we are given two numbers, N and D. Our task is to create a Program to find remainder without using modulo or % operator in C++.. If the variable a is completely divisible by the second number (b), it returns zero (0), or the remainder becomes 0. I can mail you a source snippet. This should be a compiler optimization issue. Maybe you’d like to benchmark this? Therefore, no conversion described in this article is required. Let's understand the following example that illustrates the functionality of the modulus operator. Operators that are in the same cell (there may be several rows of operators listed in a cell) are evaluated with the same precedence, in the given direction. time -= days * SECONDS_PER_DAY; /* time now contains the number of seconds in the last day */ Explain to me please how if C=A%B, C is also equal to A – B*(A/B). I do two divisions by 10 and keep both remainders. I thus sat down to write my own code. modulo operation - is a way to determine the remainder of a division operation { The modulo (%) operatorâs standard feature is to compute the remainder of a division. } In which case INT(A/B) = INT(6/4) = 1. msb=nsb=0; On an ARM M3, the div operation is good: 2 to 12 cycles depending on the data, in which case, the line I wrote: a = seconds >> 7; days = (7 +97*a + ((93*a)>>10) â((59*a)>>17) ) >>16. I thus recoded attempt 2 to use smaller data types. A useful scale factor is ceiling(2^32 * 2^n / 60), where n is the largest integer such that (2^n / 60) < 1. I thus was rather concerned about its performance and so I measured its speed for three different architectures – AVR (8 bit), MSP430 (16 bit), and ARM Cortex (32 bit). After all one is usually using an 8/16 bitter for a reason. or at least try library div/mod functions first. while (value>99) Yes we can check a number is even or odd without using division and modulus operators for that we need to use & operator; number &1 ==1 then it is an even number. Decimal to Binary using recursion and without using power operator. The increment operator is supported in two forms: the postfix increment operator, x++, and the prefix increment operator, ++x. In programming, the operator symbol tells the compiler to perform a particular operation at a given number based on the passed operation. time_sec = time_sec % 3600; To be fair, I just tested using IAR EWARM and it required medium optimization settings to generate one call to the divide library function: MOV R0,R6 Now in some cases you absolutely have to use the modulus operator. btw i don’t mean to sound rude, but i’m doing a math project on the modulus operator (which is like the coolest operator i’ve seen yet other than powers) and it’s usefulness. It shouldn’t affect execution speed, but my mind is damaged by MISRA C. I also take away from your post that if I have a decent compiler and a chip with HW multiply and divide, none of this tweaking matters in the slightest. In the next few days I’ll try and benchmark your suggestion and post the results. You are correct in that a %8 on an uint16_t is optimized away to a mask, but I think the problem is more likely to be between 8 / 16 than signed / unsigned. I’m buried with work right now Barbara. stime = time – ((uint16_t)hours * 3600U); /*stime now contains the number of seconds in the last hour */ The modulus operator finds the division with numerator by denominator which results in the remainder of the number. q = q2 + (r > 9); If rounded downwards, we might get a number of days that is one too low, but there is no risk of getting one day to many since the closest input we can get is 1 second short of a whole day. The operator takes two operands and returns the reminder after performing division of dividend by divisor. Similarly, 7 % 2 returns 1 as a remainder because when 7 is divided by 2, it returns 3 as quotient and 1 as remainder. 13 modulus 4 will be 1. I do quite a lot of this type of benchmarking and one of the problems you immediately run into is that your test code gets optimized away. However if you are working on smaller processors then clearly something needs to be doneÂ – and so I investigated some alternatives. nsd = rem[r]; The problem is that C provides to access to the underlying instructions. Thus this suggested an improvement that would make the code a lot more portable – and that is to use the C99 fast types. q1 += q1 >> 8; So (on a two’s complement machine) the compiler can only replace a modulo-power-of-2 with an logical AND if it can prove that the operand is always positive (obviously, the easiest way for the compiler to prove this is if it is an unsigned type). Modulo Operator (%) in C/C++ with Examples. I recorded the number of cycles in this second case. q1 += q1 >> 4; I then stripped out the test code and had printf() output constants. Altogether six multiplies, three subtracts, one add, and if the compiler is good, the shifts are just an address selection and do not take machine cycles at all. It has some unique properties. Example: When we perform the modulus operator between 8 and 5, means 8 % 5, it returns the remainder 3 because when 8 is divided by 5, it returns 1 as the quotient and 3 as the remainder. q3 += q3 >> 4; 4 multiplies, 2 subtracts, maybe 2 shifts if your compiler isn’t too smart. VHDL) the case is strongest for avoiding a div operation at all costs. In practice, depending where the results go, about 20 cycles on an ARM M3. { The method works by avoiding the divide operation: rather than dividing by a constant K, you multiply by a pre-calculated ciel(2^32 / K ) and then look at the more significant 32-bits of the result. }. Modulo computes a remainder. However, you can get it through a function call, div() or ldiv() which are in stdlib.h. It depends on how costly a division operation is compared to a multiply or a shift. This is not pc-specific. So we can split it into an integer and a decimal part. Operator przypisania ("="), jak sama nazwa wskazuje, przypisuje wartoÅÄ prawego argumentu lewemu, np. Before understanding about modulus operator, we need to know about the term operator. Now my question: Does this approach make any difference in the performance of the code? With a dividend of type uint8_t (instead of int, as (a+1) turned out to be), the optimizer is in fact smart enough to see that it is always positive, and does use a mask. return r; You can follow any responses to this entry through the RSS 2.0 feed. Modulo remainder is one of the two results that any reasonable processor or compiler library outputs for a division operation. This time the algorithm uses only 32-bit integers with no intermediate 64-bit results, so it can be written in straightforward C with no typecasts etc. Some microprocessors even do that automatically (BCD format: DoW, hour, min and sec take each a nibble of a 32 bit register, for example). BUT FOR A COMPLETELY different reason than the one expressed above. q2 = q1 >> 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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