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 thatâs 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 â¦ 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. Itâs 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.Â 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 Â§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.Â 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 Â§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 itâs 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

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