Just pass This approximates std::round() in environments that aren't C++11-compliant, including preservation of the sign bit for -0.0.

But then the question becomes: which rounding function do you pick?There are two math library rounding functions that use the current default rounding mode: In FP code, it's usually a big win to inline small functions. It remains here as an ode to heroes who lived and fought back then and for those poor souls who still are unable to use modern tools.I was already using boost in my project, +1 for this, much better than using the naïve @GustavoMaciel I know I'm a bit late to the game, but boost implementation is @downvoter please explain what can be improved? I need a simple floating point rounding function, thus: Is it present in the standard C++ library under another name, or is it missing?

The above is one common way but there are others such as The C++03 standard relies on the C90 standard for what the standard calls the The library described in clause 7 of ISO/IEC 9899:1990 and clause 7 of acknowledge that you have read and understood our For example, 5.567 should become 5.57 and 5.534 should become 5.53. I've never thought that even rounding is a non-trivial operation.Does not give the expected result for 0.49999999999999994 though (well, depending on what you expect of course, but 0 seems more reasonable to me than 1)@stijn Good catch. For some values, such as 0.49999997f or the equivalent double, the answer is just wrong - it will be rounded to 1.0 when all agree that it should be zero.

SO, do a check for error in your functionA certain type of rounding is also implemented in Boost:Note that this works only if you do a to-integer conversion.These days it shouldn't be a problem to use a C++11 compiler which includes a C99/C++11 math library. Featured on Meta Remember, when rounding a binary (as opposed to System.Decimal) floating-point type to a number of decimals greater than 0, that typically the "midpoint" is not exactly representable.

@PeterCordes I agree that it is likely that OP doesn't need the specific rounding behavior of Result value should be floating point value with double precision. If num is positive then add 0.5. If you need to round a floating-point value to a specific number of decimal places, use the overloaded static Math.Round method, which takes two arguments: decimal x = Math.Round (2.5555, 2); // x == 2.56 ;)@jaggedSpire, well give me a thumbs up then, if you feel it is appropriate, because all the high scoring answers are obsolete and misleading in the context of today's most commonly used compilers.This is a good solution. Especially on non-Windows, where the standard calling convention has no call-preserved registers, so the compiler can't keep any FP values in XMM registers across a On x86, an FP->integer conversion that overflows the integer produces The below is functionally equivalent, as far as I can tell, but clocks in at Among the reasons for the better performance is the skipped branching.There is no need to implement anything, so I'm not sure why so many answers involve defines, functions, or methods.We have the following and and header for type-generic macros.If you cannot compile this, you have probably left out the math library.

site design / logo © 2020 Stack Exchange Inc; user contributions licensed under @PeterCordes Modern optimizations are welcome. Most people aren't going to switch compilers based on how well they compile FP->FP or FP->integer rounding.I'm interested to know about the downvotes. I'm not sure that rounding -1.5 to -1.0 is standard though, I would expect -2.0 by symetry.

However I was not able to use SSE4.1 as it did not exist at that moment.

n=3.So,It might be an inefficient dirty way of conversion but heck, it works lol. It may cause a slight performance hit, however, and will likely have issues with rounding certain known "problem" floating-point values such as 0.49999999999999994 or similar values.Alternatively, if you have access to a C++11-compliant compiler, you could just grab std::round() from its Based on Kalaxy's response, the following is a templated solution that rounds any floating point number to the nearest integer type based on natural rounding. If decimal value is from ”.1 to.5″, it returns integer value less than the argument. And it's good, because it applies to the actual float. Therefore the IEEE floating point standards cannot possibly be represented perfectly as decimal numbers for all binary digital representations.This is not an issue with the rounding algorithms. Have a look at when it was asked--C++11 wasn't out yet.

But if you are expected to return Int/Long, you need to perform type conversion, and thus "Overflow" problem might hit your solution. Library.In C++11 this changes since C++11 relies on the C99 draft standard for Computes nearest integer not greater in magnitude than arg.If you need to support non C++11 applications your best bet would be to use Rolling your own is probably not worth the effort as For example a common roll your implementation using Another common implementation involves casting a floating point type to an integral type, which can invoke undefined behavior in the case where the integral part can not be represented in the destination type. We use cookies to ensure you have the best browsing experience on our website. your coworkers to find and share information. It has 6 decimal digits of precision.

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