modulus operator in c

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. A variable of type uint8_t should be a hint obvious enough for the optimizer to know the value is unsigned, shouldn’t it? You can leave a response, or trackback from your own site. 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).. Quite frankly, the comments (especially from those with an alternate point of view) are one of the main reasons I write this blog. In assembly language, the divide operation produces the result of the division and a remainder – the number that you want with mod. I run, use Visual Studio 2005, result print: 22 13 14 14. Modulus Operator (%) is a binary operator that’s why used with two operands. This is what PC-Lint had to say: PC-lint for C/C++ (NT) Vers. Also, I believe that the second equation given, C = A – B *(A/B), will always produce zero if you were to write it out. The variables being masked / moduloed are uint8_t. q3 = q3 >> 3; By the way, sorry for placing this comment into the wrong place. 4. The modulus operator. 24, Mar 20. In C++, the modulus operator is a percent sign, %. The results were encouraging: Although the MSP430 time increased very slightly, the AVR and ARM stayed at their fastest speeds. Lint did not find the most severe one. Hi Nigel I have come across this issue more than once. The code I found wasn’t great and as usual was highly PC centric. You can find modulus operator in almost all programming language. If you have to do it by hand it cost one comparison (besides the increments) per second, one comparison per minute, per hour, and so on… If you have to update your display once a second anyway (displaying time), you’d certainly win. The modulus … 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. Operator precedence is unaffected by operator overloading. It’s obviously more for speed and micro-optimization junkies, but you’ll see several different faster ways than the straight modulus operation. while (value>9) While I agree that 32 bit operations on small processors is painful, I don’t agree with you that this means that one should not do them. value -=100; In practice, depending where the results go, about 20 cycles on an ARM M3. NIOS), with a single cycle multiply, and no div instruction, the library div function might take several hundred cycles, you would see plenty of benefit from avoiding division even at the high cost of the line I just quoted. uint16_t q1; i made a simples test, with integers, and the results were not equal for a lot of given numbers, That is only useful when n is a power of 2…, You DEFINITELY have to be VERY carefull about using the % operator…. It helps to enhance the programmer's logical and mathematical ability by using various operators in the program. I do two divisions by 10 and keep both remainders. You can write it this way: 13 % 4 = 1. Following are the possibilities when the first number is divided by the second number to return only a remainder value. units = i-tens * 10. I thus sat down to write my own code. The assembly code the compiler produced for both code segments was identical! 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++.. The modulus operator is a symbol used in various programming languages. Consider, for instance, the following C code: int remainder = value % 1024; It can be translated into: int remainder = value & 0x3FF; In general, if divisor is a power n of two, the modulo operation can be translated to a bitwise AND with divisor-1. printf(“%d %d %d %d”,b,a++,a,++a); = Assign value from right operand to left operand. I can’t see the advantage of doing this. Writing a single piece of code for all three architectures and and expect it to give optimal performance is a little naive, IMHO. minutes = stime / 60U; r = n – q2 * 10; If you run your code under the simulator then you will almost certainly find a cycle counter. This approach has a nice looking symmetry to it.  However, it contained three divisions and three modulus operations. All rights reserved. Given two positive numbers a and n, a modulo n (abbreviated as a mod n) is the remainder of the Euclidean division of a by n, where a is the dividend and n is the divisor. The scale factor of ceiling(2^32 * 32 / 60) can be considered a 32-bit unsigned binary fraction representing (32/60). 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. b : c; because the precedence of arithmetic left shift is higher than the conditional 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. i = i- hundreds *100 }. Whenever one needs to do integer multiplications starting with an integer whose range is not limited, if the incoming number is high, multiplying it by any number greater than 1 will cause an overflow. But if you need to do a lot of time comparisons in your code, or do temporal arithmetic (e.g. Speaking of NIOS, and the FPGA world, most FPGAs have plentiful and excellent hardware multipliers. Speaking of modulos and optimizers, the IAR optimizer for MSP 430 does not always optimize modulo by powers of 2 and make them mask operations. 50. MOV R1,R7. Here’s one more idea for converting a 32-bit integer of seconds into days, hours, minutes and seconds. stime = time – ((uint16_t)hours * 3600U); /*stime now contains the number of seconds in the last hour */ 4 % 2 = 0 4 is even as remainder is 0. The optimizer does know how to use a mask instead of %, if the dividend is uint16_t, but not in this case? uint32_t days, hours, minutes, seconds; This is another example of upward rounding that makes these algorithms work. No division required! Now it’s clear for me:). : void compute_time(uint32_t time) Example: 7 % 5 = 2 Dividing 7 by 5 we get remainder 2. Depending on how often the display function runs, I’d often rather store seconds, minutes, hours, and days in 4 separate variables. Now in some cases you absolutely have to use the modulus operator. But my friend use Turbo C, the result is: 22 13 13 13. 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. The result is much slower, and actually uses more memory (2 words more, besides the DivMod16s function itself). With the constants ready at compile time, at run time we do this: I used the modulo (checking that the divisor was a power of 2 at compile time), because I took for granted that the optimizer would replace them with masks, and I felt it was more readable and logical. It uses the percentage sign character in the lexical syntax. As another example, 25 / 7 = 3 remainder 4, thus 25 % 7 = 4. Thus this suggests a simple optimization to the algorithm. “LINT: D:\temp\junk.c (6, 33) Warning 564: variable ‘a’ depends on order of evaluation” Furthermore, hours, minutes and seconds are inherently limited to an 8 bit range. You seem to be correct, thank you. ^ 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 Thanks for your insight, particularly the value of rounding up. Find the largest n where (2^n / K) < 1, then scale the result by (2^-n), which is just a right shift by n bits (for unsigned operands). This is ridiculous. { The modulus operator works with two operands received by the end-user. Modulo Operator (%) in C/C++ with Examples. 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. But were it not for the undefined behavior, the same expression would also rely on the order of evaluation, which is unspecified behavior (ISO 9899:1999 6.5 §3). I recorded the number of cycles including the printf(). http://cc.davelozinski.com/c-sharp/use-the-modulus-operator-or-alternative. 13 modulus 4 will be 1. I looked at some code generated by GCC and it appears that the divide routine produces the modulus simultaneously. shifts and ‘and’ are fast ways to div and % for powers of 2. did you know the intel pentium divide is not even used. 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 a % b gives 0 (As 10/5 will have 0 remainder) Assignment operator Assignment operator is used to assign value to a variable. It’s interesting if you don’t have a native div or mult. It can take any 32-bit, signed or unsigned. Furthermore, the uint8_t variable I apply the modulo to is incremented by a inc.w instead of inc.b. Nevermind that part. Syntax of C++ Modulus Operator […] This post was mentioned on Twitter by Michael Barr, Newsery 4. I have found that the IAR Embedded Workbench for ARM compiler does exactly this. In this tutorial, we shall learn how to use Arithmetic Modulus Operator with values of different datatypes using example programs. { – All operations that can be made on 16-bit operands are forced to be 16 bit. Developed by JavaTpoint. 3. I haven’t tried it, but I think it would be slightly more 16-bit friendly. Very interesting post Jacob. The operator takes two operands and returns the reminder after performing division of dividend by divisor. Very interesting. It is used to perform different operations (+, -, *, /) on numbers. { Only terribly unoptimized code or awful compiler optimization requires TWO divides to produce the modulo AFTER the identical division is performed. The following table shows all the arithmetic operators supported by the C language. So the register situation would be the same, but the optimizer would replace the mod to a mask. hours = (seconds * Hours_Mul) >> 32 Thank you. I think the right result is 22 13 13 13. The following code should optimize to one CPU instruction or compiler library call, if they are grouped close together in executon: time_hrs = time_sec / 3600; 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. The most critical part is to get the number of days. It’s called div. 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. As I have recently found out, not fully understanding how an operator is implemented can cause one all sorts of problems. static const uint8_t rem[12] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1}; 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. The method I describe here makes use of the fact that the M3 can multiply 32-bit integers and give the most significant 32 bits of the result. 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. In C++, Modulus is performed using arithmetic operator %. Most uCs provide a remainder after a divide in hardware and as painful as it is, I bite the bullet and simply move the remainder contents to a working register with a short assembly sequence . However examination of attempt 2 shows that further optimizations are possible by observing that if seconds is a 32 bit variable, then days can be at most a 16 bit variable. Thus, 7 divided by 3 is 2 with 1 left over. My naive implementation of the code looked like this: All I have done is change the data types and to add casts where appropriate. msd = q3 + (r > 9); I use IAR’s compilers – and all their simulators I’ve used have the feature. Computing modulus. However your point that the remainder (modulus) inherently falls out of the division operator is valid. Write code which takes platform limitations into account, of course. Arithmetic operators in C++: Operator: Description: Example + Addition of two operands: a + b will give15. ‘compute_time’ doesn’t return anything, and only uses local variables; that is, it has no side-effects at all. Again we multiply the decimal part (0.269629..) by 2^32 and round it upwards. Thus my claim that the modulus operator can be very inefficient is true for some architectures – but not all.  Thus if you are using the modulus operator on an ARM processor then it’s probably not worth worrying about. Modifying a variable twice without sequence points in between is undefined behavior (ISO 9899:1999 6.5 §2). q1 += q1 >> 4; It's important to know the difference and what the language being used uses. This entry was posted The method I describe here makes use of the fact that the M3 can multiply 32-bit integers and give the most significant 32 bits of the result. Thus the modulus becomes 6 – 4 * 1 = 2, which I believe is the correct answer. However in many cases it’s possible to restructure the code such that the modulus operator is not needed. FWIW C does have a divide with remainder function. Maybe you’d like to benchmark this? If the first number (a) is non-zero and the second number is 0, it gives an error at compile time. In this C program, we will learn how can we find the remainder of two integer numbers without using modulus (%) operator? remainder = dividend % divisor; Finally, the quotient and remainder are displayed using printf( ) . In Embedded Systems Design there is an increasing need for some form of real time clock. What kind of ‘running cycles’ statistical tools do you used? I wrote out the algorithm below in C and compiled it (with optimizations turned on), along with the usual y = x / 60 and r = x % 60 calculations. minutes = seconds * MinutesMul >> 32 So, on 2 different architectures (ARM and Intel for example), % will give different results depending on the sign of the inputs! 2. This article will introduce how to use the modulo operator in C++. Most, if not all modern CPU have hardware support for % operator (as a mater of fact, most will have a moddiv instruction which calculates division and modulo at the same time). Mail us on [email protected], to get more information about given services. Nigel Jones is an embedded systems consultant with over 20 years of experience designing electronic circuits and firmware. printf(“%d %d %d %d”,b,a++,a,++a); 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. For example most (all?) As a result it’s hardly surprising that code that uses the modulus operator can take a long time to execute. The key to this is to round UPWARDS when pre-calculating the constants. Or we can say it returns the remainder to a non-zero integer value. It is even simpler to split an integer up to 999 into hundreds, tens and units without mod or div with an ordinary 32-bit processor and no tricks. hundreds = (i*656) >> 16 The obvious way, using the modulus operator is as follows: However, using the technique espoused above, we can rewrite this much more efficiently as: If you benchmark this you should find it considerably faster than the modulus based approach. For example, std:: cout << a ? The key to this is to round UPWARDS when pre-calculating the constants. adding 5000 seconds to a time), a single 32-bit counter makes for much clearer and less buggy code, which is surely the most important consideration. (from http://people.ee.duke.edu/~dwyer/courses/ece52/Binary_to_BCD_Converter.pdf #include int main () { printf ( "%d \n " , 4 % 7 ); return 0 ; } This is a simple example of modulus operator where the remainder of the division of 4 by 7 is returned by the operator. In general I’d rather you emailed me these sorts of questions rather than put them in the forum. This leftover, or 'remainder,' is a result of a modulus operation. Therefore, no conversion described in this article is required. The most important results to remember from all this are: 1) One can easily derive integer approximations that can deliver flawless exact integer results in all cases provided that the constant divisor is small. However if you are working on smaller processors then clearly something needs to be done  – and so I investigated some alternatives. An earlier commenter had implied this in mentioning the standard div() function. Well a little thought shows that C = A % B is equivalent to C = A – B * (A / B). Thanks for the hint on usual arithmetic conversions. Well I found myself in just such a situation recently. Modulus Operator % It is used to find remainder, check whether a number is even or odd etc. 9.00f, Copyright Gimpel Software 1985-2010, — Module: D:\temp\junk.c (C) In programming, the operator symbol tells the compiler to perform a particular operation at a given number based on the passed operation. Each programming language has same usage of modulus operator and in C++ this operator is also called as remainder operator. The second line has an unexpected start: a constant 7 is added. c needs a divide-with-remainder like every cpu has. Prepare to multiply by 2^32 / (24 * 60*60) which we can pre-calculate. Things are different in C++ where it was left up to the implementation, but a similar standard to C99 was recommended as of C++98 and it has since been defined in C++11. 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. Operator przypisania ("="), jak sama nazwa wskazuje, przypisuje wartość prawego argumentu lewemu, np. uint32_t r = x – (q * 60u); This is a list of operators in the C and C++ programming languages.All the operators listed exist in C++; the fourth column "Included in C", states whether an operator is also present in C. Note that C does not support operator overloading.. I think this is only true for some 32 bit assemblers. ^ 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. Line 6: 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 from attempt 1 to attempt 2 you change the orientation of your code. –. It would probably result in mov.w R15,R15 (unnecessary since R15 is already written to by a mov.b operation, which zeroes the high byte). Newsery 4 said: Efficient C Tip — use the modulus (%) operator with caution – http://bit.ly/eAtNJ2 – [Hacker News FH] […]. value -=10; The second operator yields 1. Clearly the ARM doesn’t like working with non 32 bit variables. Decimal to Binary using recursion and without using power operator. “LINT: D:\temp\junk.c (6, 37) Warning 564: variable ‘a’ depends on order of evaluation”, Thank you, Mr Nigel Jones. For example, the expression a = b = c is parsed as a = (b = c), and not as (a = b) = c because of right-to-left associativity. Thus although I have replaced a single operator (%) with two operations (- *) I still expect an increase in speed because the modulus operator is actually three operators in one (- * /).  Thus effectively I have eliminated three divisions and so I expected a significant improvement in speed. sorry: I do not know of a “C” compiler that provides access to the remainder. Modulo remainder is one of the two results that any reasonable processor or compiler library outputs for a division operation. JavaTpoint offers college campus training on Core Java, Advance Java, .Net, Android, Hadoop, PHP, Web Technology and Python. In the C Programming Language, the fmod function returns the remainder when x is divided by y. Thus attempt #4 is both fast and portable. q2 = q1 >> 3; And if not, why? This is not pc-specific. If the variable a is not completely divisible by the second number (b), it returns a numeric value in the range [1, a - 1]. void my_display_value(uint16_t n) The standard div() function is intended to allow the compiler to produce the quotient and remainder from a single division operation. In all three cases I used an IAR compiler with full speed optimization. MOV R0,R1 What do you think of these changes? Now my question: Does this approach make any difference in the performance of the code? MOV R1,#+3600 ‘David. I recorded the number of cycles in this second case. Jacob is correct to point out that the division by 60 or modulo 60 operation can also be transformed into a multiplication by the reciprocal of 60 with scaling. C Program to find the roots of quadratic equation, How to run a C program in Visual Studio Code. As far as I know, there are instruction set extensions with division for some members of different MCU families. b = a++ + ++a; 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. If you insist on conversion approach, and still care about speed, you would probably need a custom division code, optimized for dividing 32-bit uint by an unsigned byte. Program to find remainder when large number is divided by 11. modulus sometimes come for free if the two operators (% and /) are in the same block in the source and use the same operands, . In other words the modulus operator is functionally equivalent to three operations. Anyone with time on their hands care to benchmark this lot versus the mod and div approaches? Submitted by IncludeHelp, on April 26, 2018 . As a result, I thought a quick internet search was in order to find the ‘best’ way of converting ‘time’ to days, hours, minutes and seconds. and that is a VERY hard bug to find in your code! For 8 and 16 bit MCUs using 32-bit math is wrong approach altogether, and benchmarks show it quite clearly, no matter what “optimizations” you try to apply. The modulus operator in C is denoted by % (percentile) operator. Jacob: You might find this posting interesting. You can also scale the factor (2^32 / K) up by a power of 2 for a more accurate result with larger divisors. lsd = rem[r]; q3 = (q >> 1) + (q >> 2); It depends on how costly a division operation is compared to a multiply or a shift. on Tuesday, February 8th, 2011 at 9:21 am and is filed under Algorithms, Efficient C/C++, General C issues. I would think it also makes for smaller and faster object code too, though I don’t currently have the means to check that. In programming, the operator symbol tells the compiler to perform a particular operation at a given number based on the passed operation. We can use the Chaining of Modulus Operator to perform modular division on more than two operands in a single statement. Modulus Operator in C/C++. The % (modulo operator) is an example of such an operator, as it is not defined the same way in all programming languages. Now we have DaysIntMul = 49710 and DaysDecMul = 1158050442, Similarly we prepare HoursMul =ceil(2^32 / (60*60) ) = 1193047 Seconds = seconds – days * SecondsInDay If it manages to optimize on uint16_t, and makes the operations on uint8_t as if they were on uint16_t, why doesn’t it manage to optimize the uint8_t case? Is it right? seconds = seconds – 60* minutes. I am using Keil MDK for Cortex, but I never found there is such a tools. The C language does not support this directly, but my compiler will generate good code if you declare a local 64-bit integer and immediately right-shift it. © Copyright 2011-2018 www.javatpoint.com. Casting to (unsigned) does work. seconds = seconds – hours * SecondsInHour We need to effectively divide by (24*60*60). bro, you know that mod is very efficient for all operands of powers of 2 due to optimized instructions for this special binary case. Consider a division by 60 or modulo 60 operation. I wrote that in my comment, but since it was between less-than and greater-than signs, it disappeared…. Again, note how this works correctly only when the constants are rounded upwards. This is not very readable but should be pretty fast! I always configure my compilers to give the maximal amount of warnings. Modulo operation - Wikipedia, the free encyclopedia. Thank you mr. Nigel. Thus an output of “My brain hurts” would in fact be ‘correct’. { or at least try library div/mod functions first.

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