floating-point-precision

This question already has an answer here: Why do I see a double variable initialized to some value like 21.4 as 21.399999618530273? 14 answers I am wondering why I am losing precision when using this code : double x = 12.0456; // or float : same result System.out.println(x); // outputs 12.0456 obviously x %= 1; // should now be equal to 0.0456 right? System.out.println(x); // outputs 0.04560000000000031 or 0.045599937 when using float 12.0456 modulo 1 should equal 0.0456 right? But it shows a slightly different value, why do I keep losing precision? I mean the code should substract exactly 1

This question already has an answer here: Comparing float and double 3 answers According to this post , when comparing a float and a double, the float should be treated as double. The following program, does not seem to follow this statement. The behaviour looks quite unpredictable. Here is my program: void main(void) { double a = 1.1; // 1.5 float b = 1.1; // 1.5 printf("%X %X\n", a, b); if ( a == b) cout << "success " <<endl; else cout << "fail" <<endl; } When I run the following program, I get "fail" displayed. However, when I change a and b to 1.5, it displays "success". I have also

Are we safe to use floats as loop-counters and to increment/decrement them by fractional amounts at each iteration,like in the seemingly risk-free program below?Of course I know that using floats as operands for the == operator is a dumb thing to do.But what's wrong with using floats as operands for other comparison operations for "normal" purposes? By "normal" I mean that,well,even though floats may not be the exact numerical representation of the number,but isn't a variation like 0.000000001 irrelevant and can be ignored in most cases? (For example in the following program that isn't even

When I am using (int) with (double) some times it is not working correct. Look At The PHP Code Example: I Need To LEAVE 2 Decimals And REMOVE Other... I Know number_format(); function But I Cannot Use It. Because It Is Rounding Number number_format(24.299,2); Output: 24.30 I Need: 24.29 <?php $str="158.2"; echo (double)$str; // Output: 158.2 echo (double)$str*100; // Output: 15820 echo (int)((double)$str*100); // Output: 15819 <-WHY? It Must To Be 15820, Why 15819? echo ((int)((double)$str*100)/100); // Output: 158.19 ?> I need To leave two decimals in the number and cut other WITHOUT rounding

Where should I look at if I want to switch between float and double precision at compile time. Its like, if user wants everything in float instead of double precision how I can maintain this flexibility? In other words, how should I define a variable that could be either float or double precision conditionally? If it is OK to make the switch at compile time, a simple typedef would do: #ifdef USE_DOUBLES typedef double user_data_t; #else typedef float user_data_t; #endif Use user_data_t in your code, and set USE_DOUBLES if you want doubles at compile time: g++ -DUSE_DOUBLES=1 myprogram.cpp

I am trying to affect the translation of a 3D model using some UI buttons to shift the position by 0.1 or -0.1. My model position is a three dimensional float so simply adding 0.1f to one of the values causes obvious rounding errors. While I can use something like BigDecimal to retain precision, I still have to convert it from a float and back to a float at the end and it always results in silly numbers that are making my UI look like a mess. I could just pretty the displayed values but the rounding errors will only get worse with more editing and they make my save files rather hard to read.