Why is this a narrowing conversion? Because the definition includes (with my emphasis): C++11 8.5.4/7 A narrowing conversion is an implicit conversion […] from an integer type […] to a floating-point type, except where the source is a constant expression and the actual value after conversion will fit into the target type and will produce … Read more
Firstly, a Java double cannot be a Java null, and cannot be compared with null. (The double type is a primitive (non-reference) type and primitive types cannot be null.) I’m therefore assuming the following: The “null” you are trying to detect is a NULL stored in the database that you are querying. You are using … Read more
yourTextView.setText(String.format(“Value of a: %.2f”, a));
Because Convert.ToInt32 rounds: Return Value: rounded to the nearest 32-bit signed integer. If value is halfway between two whole numbers, the even number is returned; that is, 4.5 is converted to 4, and 5.5 is converted to 6. …while the cast truncates: When you convert from a double or float value to an integral type, … Read more
You’re performing an integer division and then type casting: tf = (double) ( word.getValue() / noOfTerms ); ^—–integer division—-^ Type cast one of the elements in the division to convert into a floating point division: tf = ((double)word.getValue()) / noOfTerms;
how did Walter Laan says in his thread Never give up! Never surrender! EDIT: I can’t resist, but due to my poor English I dare not to commenting why, where and how is that possible, nor works correctly, for confirmations I added Rob’s two (little bit) modified class for TableColumnRendering …, import java.awt.EventQueue; import java.math.RoundingMode; … Read more
Check answer here. You can write moneyToGet.setText(String.format(“%.0f”, priceValue));
Comparing floating point values using the == operator is very error prone; two values that should be equal may not be due to arithmetic rounding errors. The common way to compare these is to use an epsilon: bool double_equals(double a, double b, double epsilon = 0.001) { return std::abs(a – b) < epsilon; }
I don’t understand why all the answers to that question are the same. There are several ways to calculate the square root from a number. One of them was proposed by Isaac Newton. I’ll only write one of the simplest implementations of this method. I use it to improve the accuracy of double’s square root. … Read more
That’s because integers don’t have values for +/-Inf, NaN, and don’t allow division by 0, while floats do have those special values.