I want to remove the decimal point from double?
Ex: 234.567 will become 234567.
You can use String.valueOf method and then replace . with a blank character("").
String s = String.valueOf(234.567).replace(".","");
Use a easy way out.
Cast it to int. :P
x = (float) 5.12;
int a = (int) x;
System.out.println(a);
Related
Hello new to android/Java,
I am using JSON to Parse values as strings and doubles. I am getting strings/doubles such as "6503.04" or "12.3942" etc. I am looking to see if I can convert these strings into an integer and also doubles into integers. I just need to get rid of the decimals points in the easiest way possible. How can I make that happen?
Any help appreciated.
double x = Double.parseDouble(your_string);
int y = (int) x;
y is going to have value of x with decimals cutted of.
Before casting to int you are able to floor or ceil the number if you want.
Don't forget that parsing functions usually throw an exception when your_string is not a number.
//get the value from json as double then get the integer
Double number = jsonObject.getDouble('longitude')
int newNumber = number.intValue()
The Java code is as follows:
String s = "0.01";
int i = Integer.parseInt(s);
However this is throwing a NumberFormatException... What could be going wrong?
String s = "0.01";
double d = Double.parseDouble(s);
int i = (int) d;
The reason for the exception is that an integer does not hold rational numbers (= basically fractions). So, trying to parse 0.3 to a int is nonsense.
A double or a float datatype can hold rational numbers.
The way Java casts a double to an int is done by removing the part after the decimal separator by rounding towards zero.
int i = (int) 0.9999;
i will be zero.
0.01 is not an integer (whole number), so you of course can't parse it as one. Use Double.parseDouble or Float.parseFloat instead.
Use,
String s="0.01";
int i= new Double(s).intValue();
String s="0.01";
int i = Double.valueOf(s).intValue();
This kind of conversion is actually suprisingly unintuitive in Java
Take for example a following string : "100.00"
C : a simple standard library function at least since 1971 (Where did the name `atoi` come from?)
int i = atoi(decimalstring);
Java : mandatory passage by Double (or Float) parse, followed by a cast
int i = (int)Double.parseDouble(decimalstring);
Java sure has some oddities up it's sleeve
Using BigDecimal to get rounding:
String s1="0.01";
int i1 = new BigDecimal(s1).setScale(0, RoundingMode.HALF_UP).intValueExact();
String s2="0.5";
int i2 = new BigDecimal(s2).setScale(0, RoundingMode.HALF_UP).intValueExact();
suppose we take a integer in string.
String s="100";
int i=Integer.parseInt(s);
or
int i=Integer.valueOf(s);
but in your question the number you are trying to do the change is the whole number
String s="10.00";
double d=Double.parseDouble(s);
int i=(int)d;
This way you get the answer of the value which you are trying to get it.
use this one
int number = (int) Double.parseDouble(s);
Use Double.parseDouble(String a) what you are looking for is not an integer as it is not a whole number.
One more solution is possible.
int number = Integer.parseInt(new DecimalFormat("#").format(decimalNumber))
Example:
Integer.parseInt(new DecimalFormat("#").format(Double.parseDouble("010.021")))
Output
10
I am newbie in Java and just trying to convert a variable according to math operation. Let's have following scheme:
int var_result;
if(addition) {
var_result = number1+number2; // number1, number2 => integers
}
else if(division){
var_result = (float) number1 / number2;
// and here is the problem - I don't know, how to convert "var_result" from integer to float
}
...just print var result ...
Is there any quick way, how to convert the var_result variable in this case? The easiest way in this example would be don't convert it and just use float var_result_float, but I don't want to solve it this way...
Thanks
int var_result; is already declared. You can't expect to store a float in it, without losing precision.
When you perform type-casting, the variable is temporarily treated as another type for the purpose of evaluation of an expression but that does not change what it actually is.
An int remains an int. You should go with float var_result_float to store a float.
int var_result is an int variable.
If you need a float use a float var_result
I don't want to solve it this way...
Well...This is the only way.
If you need to store an int or float in a variable you can use a double to store all possible values.
IMHO Don't use a float if you can possibly avoid it as its precision is fairly poor.
You can't change type of variable, but you can get what you want this way:
Object var_result;
if(addition) {
var_result = new Integer(number1+number2); // number1, number2 => integers
}
else if(division){
var_result = new Float( (float) number1 / (float) number2);
}
Of course to use the number, you have to test the type with instanceof and do cast to Float or Integer as needed, so this is very clumsy.
how to convert Float to Integer in java?
Float value = 30.0F
how to convert above value to Integer?
Please help me?
Use Float.intValue():
Integer i = value.intValue();
Note that this causes autoboxing, but since you're planning to create an Integer anyway, this won't have any performance impact.
Note also that you should pay attention to rounding: intValue() and an int cast round toward zero. For rounding to the nearest integer, use Math.round(), for rounding down use Math.floor(), for rounding up use Math.ceil(). If you need some other kind of rounding, you need to implement it yourself.
Try this:
Float f = new Float(10.5);
Integer i = new Integer((int)Math.ceil(f));
f.intValue() is the way to go..
new Float(value).intValue() or simly cast it to int int v = (int) value
You can just do this:
Float value = 30.0f;
Integer intVal = value.intValue(); // auto-boxing happens here
Use value.intValue() method.
Float value = 30.0F;
Integer intValue=Integer.valueOf(value.intValue());
The Java code is as follows:
String s = "0.01";
int i = Integer.parseInt(s);
However this is throwing a NumberFormatException... What could be going wrong?
String s = "0.01";
double d = Double.parseDouble(s);
int i = (int) d;
The reason for the exception is that an integer does not hold rational numbers (= basically fractions). So, trying to parse 0.3 to a int is nonsense.
A double or a float datatype can hold rational numbers.
The way Java casts a double to an int is done by removing the part after the decimal separator by rounding towards zero.
int i = (int) 0.9999;
i will be zero.
0.01 is not an integer (whole number), so you of course can't parse it as one. Use Double.parseDouble or Float.parseFloat instead.
Use,
String s="0.01";
int i= new Double(s).intValue();
String s="0.01";
int i = Double.valueOf(s).intValue();
This kind of conversion is actually suprisingly unintuitive in Java
Take for example a following string : "100.00"
C : a simple standard library function at least since 1971 (Where did the name `atoi` come from?)
int i = atoi(decimalstring);
Java : mandatory passage by Double (or Float) parse, followed by a cast
int i = (int)Double.parseDouble(decimalstring);
Java sure has some oddities up it's sleeve
Using BigDecimal to get rounding:
String s1="0.01";
int i1 = new BigDecimal(s1).setScale(0, RoundingMode.HALF_UP).intValueExact();
String s2="0.5";
int i2 = new BigDecimal(s2).setScale(0, RoundingMode.HALF_UP).intValueExact();
suppose we take a integer in string.
String s="100";
int i=Integer.parseInt(s);
or
int i=Integer.valueOf(s);
but in your question the number you are trying to do the change is the whole number
String s="10.00";
double d=Double.parseDouble(s);
int i=(int)d;
This way you get the answer of the value which you are trying to get it.
use this one
int number = (int) Double.parseDouble(s);
Use Double.parseDouble(String a) what you are looking for is not an integer as it is not a whole number.
One more solution is possible.
int number = Integer.parseInt(new DecimalFormat("#").format(decimalNumber))
Example:
Integer.parseInt(new DecimalFormat("#").format(Double.parseDouble("010.021")))
Output
10