The biggest/largest integer that can be stored in a double without losing precision is the same as the largest possible value of a double. That is, DBL_MAX or approximately 1.8 × 10308 (if your double is an IEEE 754 64-bit double). It’s an integer. It’s represented exactly. What more do you want? Go on, ask me what … Read more
If the idea is to print integers stored as doubles as if they are integers, and otherwise print the doubles with the minimum necessary precision: public static String fmt(double d) { if(d == (long) d) return String.format(“%d”,(long)d); else return String.format(“%s”,d); } Produces: 232 0.18 1237875192 4.58 0 1.2345 And does not rely on string manipulation.
This is because you are using the integer division version of operator/, which takes 2 ints and returns an int. In order to use the double version, which returns a double, at least one of the ints must be explicitly casted to a double. c = a/(double)b;
Java prevent E notation in a double: Five different ways to convert a double to a normal number: import java.math.BigDecimal; import java.text.DecimalFormat; public class Runner { public static void main(String[] args) { double myvalue = 0.00000021d; //Option 1 Print bare double. System.out.println(myvalue); //Option2, use decimalFormat. DecimalFormat df = new DecimalFormat(“#”); df.setMaximumFractionDigits(8); System.out.println(df.format(myvalue)); //Option 3, use … Read more
Here’s an utility that rounds (instead of truncating) a double to specified number of decimal places. For example: round(200.3456, 2); // returns 200.35 Original version; watch out with this public static double round(double value, int places) { if (places < 0) throw new IllegalArgumentException(); long factor = (long) Math.pow(10, places); value = value * factor; … Read more
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
float and double are floating binary point types. In other words, they represent a number like this: 10001.10010110011 The binary number and the location of the binary point are both encoded within the value. decimal is a floating decimal point type. In other words, they represent a number like this: 12345.65789 Again, the number and … Read more
You have to convert both values to BigDecimal and then you can compare it: BigInteger bi = new BigInteger(“1”); BigDecimal db = new BigDecimal(bi); db.compareTo(new BigDecimal(1.3d));
I think you mean this.stack = new double[default_capacity]; and also this.stack = new double[capacity];
#include <stdio.h> #include <string.h> typedef struct dbl9 { double array[9]; } Dbl9; Dbl9 fun(double *in){ Dbl9 ret; memcpy(ret.array, in, sizeof(ret.array));//There can be no assurance that <in> the correct return ret; } int main(){ double array[9] = {1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0}; Dbl9 dbl9= fun(array); int i; for(i=0;i<9;++i){ printf(“%f\n”, dbl9.array[i]); } return … Read more