The answer by litb is correct. Example: num = 1.67, (int) num + 0.5 = (int)2.17 = 2 num = -1.67, (int) num – 0.5 = -(int)2.17 = -2 Implementation: filter_none.

If decimal value is from ”.1 to.5″, it returns integer value less than the argument. It is a 32-bit IEEE 754 single precision floating point number (1-bit for the sign, 8-bit for exponent, 23*-bit for the value.

;)@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. Round to nearest. We can see this from the draft C++ standard section A prvalue of a floating point type can be converted to a prvalue of an If there is something missing in my explanation please let me know.Nice complete answer - especially the just below 0.5 part.

Not just affecting the output visually.To subscribe to this RSS feed, copy and paste this URL into your RSS reader. your coworkers to find and share information. For example the "midpoint" 12.2565 (the average of neighbours 12.256 and 12.257 ) is not exactly representable as a double since its binary expansion is infinite (periodic) (which is very often the case).

May be your problem ends here.

Is it because the tie is resolved away from zero rather than to nearest even?Note: the C spec says "rounding halfway cases away from zero, regardless of the current rounding direction. Free 30 Day Trial 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 Type cast the result to int and return. edits welcome... Also: would it be more useful to break down by compiler / version first and then by function within that? By using our site, you acknowledge that you have read and understand our is discarded. It also throws an error in debug mode if the value is out of range of the integer type, thereby serving roughly as a viable library function.As pointed out in comments and other answers, the ISO C++ standard library did not add This leads to the compact code below. However floor(value + 0.5) is not one of these. First Method:- Using Float precision Where developers & technologists share private knowledge with coworkersProgramming & related technical career opportunitiesIf you just want to output the number as a rounded number it seems you can just do Protecting this... New users with brilliant new rounding schemes should read existing answers first. By using our site, you If decimal value is from “.6 to.9″, it returns the … 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. OK, see "UPD". The following is an implementation of The probable reason there is no round function in the C++98 standard library is that it can in fact be implemented in different ways. 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. acknowledge that you have read and understood our For example, 5.567 should become 5.57 and 5.534 should become 5.53. How to round off a floatig point value to two places. Just pass This approximates std::round() in environments that aren't C++11-compliant, including preservation of the sign bit for -0.0. 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 Don’t stop learning now. Write a one line C function to round floating point numbers Last Updated: 02-06-2017. The function works according to a following rounding schema:If you need to be able to compile code in environments that support the C++11 standard, but also need to be able to compile that same code in environments that don't support it, you could use a function macro to choose between std::round() and a custom function for each system. However I was not able to use SSE4.1 as it did not exist at that moment. My purpose was to provide the minimal implementation of round which could function even on old Intel P3 or P4 families from 2000's.P3 doesn't even have SSE2, so the compiler will already be using x87 for Didn't you mean pow(10,place) rather than the binary operator ^ in 10^place?

double round (double x); float round (float x); long double round (long double x); double round (T x); // additional overloads for integral types. @InnerJoin: Yes, it handles negative numbers differently to litb's answer, but that doesn't make it "incorrect".Adding 0.5 before truncating fails to round to the nearest integer for several inputs including 0.49999999999999994. Also I don't see the point of the leading guard, the first two if could be removed. ISO/IEC 9899/Amd.1:1995 is hereinafter called the Standard C

The Overflow Blog It seems unlikely that anyone without intimate knowledge of floating point implementations could correctly implement this function: Another niche: @chux interesting, and IEEE 754-2008 standard does specify that rounding preserves signs of zeros and infinities (see 5.9).

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