As others have mentioned, you’ll probably want to use the BigDecimal class, if you want to have an exact representation of 11.4. Now, a little explanation into why this is happening: The float and double primitive types in Java are floating point numbers, where the number is stored as a binary representation of a fraction … Read more
You can set the precision directly on std::cout and use the std::fixed format specifier. double d = 3.14159265358979; cout.precision(17); cout << “Pi: ” << fixed << d << endl; You can #include <limits> to get the maximum precision of a float or double. #include <limits> typedef std::numeric_limits< double > dbl; double d = 3.14159265358979; cout.precision(dbl::max_digits10); … Read more
The problem you’re having relates to how floating-point numbers are represented on a computer. A more detailed discussion of floating-point representations appears towards the end of my answer (The “Floating-point representation” section). The TL;DR version: because computers have finite amounts of memory, numbers can only be represented with finite precision. Thus, the accuracy of floating-point … Read more
In most programming languages, floating point numbers are represented a lot like scientific notation: with an exponent and a mantissa (also called the significand). A very simple number, say 9.2, is actually this fraction: 5179139571476070 * 2 -49 Where the exponent is -49 and the mantissa is 5179139571476070. The reason it is impossible to represent … Read more
Your code is equivalent to: double x1 = 0.9; float xf = (float)x1; // discard some precision double x2 = (double)xf; if (x2 >= x1) { // … } The problem is that, after you discarded some precision from x1 by converting it to float, it can never be recovered; casting it back to double … Read more
The problem is that the numbers are being calculated to a higher precision. Try, =ABS(ROUND(K2, 2)*(ROUND(K11, 2)-ROUND(K10, 2))/ROUND(K13, 2)) Look into the ‘Precision as displayed’ option but be sure to understand all of the caveats before deciding to do it.
There is nothing unreliable about numpy’s behavior in the examples you show as compared to MATLAB, nor do any of the examples you show have anything to do with floating-point issues (with one exception). For the rounding behavior, MATLAB is the one doing it wrong here. Numpy is following the IEEE standard for rounding. The … Read more