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Java Format string to C#

Don't understand Java's syntax. How to convert: "%6d %7.1f %5.1f" to C# equivalent ?
I keep getting this print out in C#: %6d %7.1f %5.1f
Tried:
"{0:d6} {7:1f} {5:1f}"
But, ran into an exception.
Exception:
Unhandled Exception: System.FormatException: Index (zero based) must be greater than or equal to zero and less than the size of the argument list.
at System.Text.StringBuilder.AppendFormatHelper(IFormatProvider provider, String format, ParamsArray args)
at System.String.FormatHelper(IFormatProvider provider, String format, ParamsArray args)
at System.String.Format(String format, Object arg0, Object arg1, Object arg2)
at experiment.Main(String[] args)
The Java code:
String.format("%6d %7.1f %5.1f", int, double, double/double);
It's obvious what values will be generated based on variable data types.
EDIT: I just looked at, Convert this line of Java code to C# code
C#
String.Format("{0:x2}", arrayOfByte[i]);
Java
String.format("%02x", arrayOfByte[i]);
PLEASE. PLEASE. PLEASE. DO not close this. Kindly. Please.

NOTE: Completely rewrote my original answer based on a (hopefully) better understanding of the Java format specifiers.
Based on my (Google-limited understanding), %6d, %7.1f and %5.1f correspond to the following:
An integer with up to 6 characters, padded if less than 6.
A float with up to 7 characters (including the decimal point and decimal portion) with a precision of 1.
A float with up to 5 characters (including the decimal point and decimal portion) with a precision of 1.
You can accomplish this with C#'s String.Format, like this:
var newString = String.Format("{0,6:d} {1,7:f1}, {2,5:f1}", 605, 20.5, 8.22);
This will result in the following string:
" 605 20.5 8.22"
The first digit in each placeholder group (defined by { and }) corresponds to the argument passed in after the string:
0 = 605
1 = 20.5
2 = 8.22
The second digit, after the , refers to the length of the string (including decimal points and decimal portions).
6 = 6 characters for the integer
7 = 7 characters for the float
5 = 5 characters for the float
The letters and numbers after the : are the format specifiers.
d = integer
f1 = floating with a precision of 1.
Which produces the string above, as follows:
{0,6:d} turns 605 into " 605" (3 leading spaces due to the 6 before the :)
{1,7:f1} turns 20.5 into " 20.5" (3 leading spaces due to the 7 before the :)
{2,5:f1} turns 8.22 into " 8.2" (1 leading space due to the 5 before the : and 1 decimal number due to the precision).
As I said earlier, check String.Format and Standard Numeric Format Strings for more information.

Starting from C# 6. you can use interpolation.
For your case you may wanted to try the following:
string formattedString = $"{0:d6} {7.1:f} {5.1:f}";
before C# 6 you can try the following:
string formattedString = String.Format("{0:d6} {1:f} {2:f}", 0, 7.1, 5.1);

How to change a float into a String in Java? [duplicate]

This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 7 years ago.
Is it possible to change a float value to a String? If possible, is it also possible to converting it to a String while rounding the number to the nearest integer?
For example if I have a float such as 2.335 then, can I change it to a String of value "2.335" or "2" (by rounding it)?

Use java Float class:
String s = Float.toString(25.0f);
if you want to round down a number, simply use the Math.floor() function.
float f = 2.9999f;
String s = Float.toString(Math.floor(f));//rounds the number to 2 and converts to String
first line rounds the number down to the nearest integer and the second line converts it to a string.
Another way of doing this is using the String.valueOf(floatNumber);
float amount=100.00f;
String strAmount=String.valueOf(amount);

To do this you can simply do
float example = 2.335
String s = String.valueOf(Math.round(example));

To convert a float to a String:
String s = Float.toString(2.335f);
Rounding can be done via
String.format("%.5g%n", 0.912385);
which returns 0.91239
For a more elaborate answer, see Round a number in Java

Your first requirement can be fullfilled with String.valueOf
float f = 1.6f;
String str = String.valueOf(f);
For roundoff you can use Math.round and not Math.floor. As Math.floor will convert 1.6 to 1.0 and not 2.0 while Math.round will roundoff your number to nearest integer.

Convert a specific double to string [duplicate]

This question already has answers here:
How to nicely format floating numbers to string without unnecessary decimal 0's
(29 answers)
Closed 7 years ago.
If I have a double storing 10.0 what is the best way to convert it to string 10?
Currently I do:
if (object.getValue() > 9.999) {
someObject.setText(String.format("%d", new Double(object.getValue()).intValue()));
}
else {
//use amount as is
}
Update
These are class grade scores. I only care for a good coding style for checking out if a double is 10.0 and only then convert it to the string “10”. Don’t care about any other number

If that is all you want to do you can do like this:
double number = 10.0;
String output = ("" + (int)number);
System.out.println(output);

The easiest way will be by doing this.
double d = 10.0
String str = String.valueOf(d);
System.out.println(str.substring(0,str.indexOf(".")));
Since we are dealing with double, rounding off can be done. Refer this SO

Try this. I think it should work.
String s=String.valueOf(10.0);
s=s.replace(".0","");
System.out.println(s);

try this:
double d = 10;
System.out.println(new BigDecimal(d).setScale(0, RoundingMode.HALF_UP).toPlainString());
will print only 10 without trailing zeros, or you can change the scale if you wish.
The advantage over casting to int is that, there is rounding, and 9.9 will be rounded to 10, while cast to int will give 9.

If you know the string, you can simply substring it until the dot:
String s = "10.0";
System.out.println(s.substring(0, s.indexOf('.')));
If you want to manipulate it as a double (I don't really see why, though), you can try something like:
System.out.println(String.format("%d", (int)Math.floor(new Double("10.0"))));
There are plenty of ways, of course. I think overall your best bet is with a substring, but I'm not sure what your needs are.
Hope this helps :)
UPDATE after question clarification:
If all you're interested is checking the value of a double (assuming of course, object.getValue() is a double), just check:
if (object.getValue() >= 10)
It will cover whatever you need and you can then print a simple "10" string if needed.

You could compare with double 10.00 as in:
final double TEN_DOUBLE = 10.00;
double ten = 10.00;
double nineAndSomething = 9.10;
System.out.println("Formatted:" + ((ten == TEN_DOUBLE) ? "10" : ten));
System.out.println("Formatted:" + ((nineAndSomething == TEN_DOUBLE) ? "10" : nineAndSomething));
This gives you the following output:
Formatted:10
Formatted:9.1

Double value with specific precision in java

I'm programming a simple java program. I need to get a string from input and divide it into two parts: 1-double 2-string.
Then I need to do a simple calculation on the double and send the result to the output with specific precision(4). It works fine, but there is a problem when the input is 0, then it doesn't work properly.
For example for these input, output will be:
1 kg
output:2.2046
3.1 kg
output:6.8343
But when the input is 0, the output should be 0.0000, but it shows 0.0 .
What should I do to force it to show 0.0000?
I read similar post about double precision, they suggest something like BigDecimal class, but I can't use them in this case,
my code for doing this is:
line=input.nextLine();
array=line.split(" ");
value=Double.parseDouble(array[0]);
type=array[1];
value =value*2.2046;
String s = String.format("%.4f", value);
value = Double.parseDouble(s);
System.out.print(value+" kg\n");

DecimalFormat will allow you to define how many digits you want to display. A '0' will force an output of digits even if the value is zero, whereas a '#' will omit zeros.
System.out.print(new DecimalFormat("#0.0000").format(value)+" kg\n"); should to the trick.
See the documentation
Note: if used frequently, for performance reasons you should instantiate the formatter only once and store the reference: final DecimalFormat df = new DecimalFormat("#0.0000");. Then use df.format(value).

add this instance of DecimalFormat to the top of your method:
DecimalFormat four = new DecimalFormat("#0.0000"); // will round and display the number to four decimal places. No more, no less.
// the four zeros after the decimal point above specify how many decimal places to be accurate to.
// the zero to the left of the decimal place above makes it so that numbers that start with "0." will display "0.____" vs just ".____" If you don't want the "0.", replace that 0 to the left of the decimal point with "#"
then, call the instance "four" and pass your double value when displaying:
double value = 0;
System.out.print(four.format(value) + " kg/n"); // displays 0.0000

System.out.format("%.4f kg\n", 0.0d) prints '0.0000 kg'

I suggest you to use the BigDecimal class for calculating with floating point values. You will be able to control the precision of the floating point arithmetic. But back to the topic :)
You could use the following:
static void test(String stringVal) {
final BigDecimal value = new BigDecimal(stringVal).multiply(new BigDecimal("2.2046"));
DecimalFormat df = new DecimalFormat();
df.setMaximumFractionDigits(4);
df.setMinimumFractionDigits(4);
System.out.println(df.format(value) + " kg\n");
}
public static void main(String[] args) {
test("0");
test("1");
test("3.1");
}
will give you the following output:
0,0000 kg
2,2046 kg
6,8343 kg

String.format is just makign a String representation of the floating point value. If it doesnt provide a flag for a minimum precision, then just pad the end of the string with zeros.

Use DecimalFormat to format your double value to fixed precision string output.
DecimalFormat is a concrete subclass of NumberFormat that formats
decimal numbers. It has a variety of features designed to make it
possible to parse and format numbers in any locale, including support
for Western, Arabic, and Indic digits. It also supports different
kinds of numbers, including integers (123), fixed-point numbers
(123.4), scientific notation (1.23E4), percentages (12%), and currency
amounts ($123). All of these can be localized.
Example -
System.out.print(new DecimalFormat("##.##").format(value)+" kg\n");

How can I format a String number to have commas and round?

What is the best way to format the following number that is given to me as a String?
String number = "1000500000.574" //assume my value will always be a String
I want this to be a String with the value: 1,000,500,000.57
How can I format it as such?

You might want to look at the DecimalFormat class; it supports different locales (eg: in some countries that would get formatted as 1.000.500.000,57 instead).
You also need to convert that string into a number, this can be done with:
double amount = Double.parseDouble(number);
Code sample:
String number = "1000500000.574";
double amount = Double.parseDouble(number);
DecimalFormat formatter = new DecimalFormat("#,###.00");
System.out.println(formatter.format(amount));

This can also be accomplished using String.format(), which may be easier and/or more flexible if you are formatting multiple numbers in one string.
String number = "1000500000.574";
Double numParsed = Double.parseDouble(number);
System.out.println(String.format("The input number is: %,.2f", numParsed));
// Or
String numString = String.format("%,.2f", numParsed);
For the format string "%,.2f" - "," means separate digit groups with commas, and ".2" means round to two places after the decimal.
For reference on other formatting options, see https://docs.oracle.com/javase/tutorial/java/data/numberformat.html

Given this is the number one Google result for format number commas java, here's an answer that works for people who are working with whole numbers and don't care about decimals.
String.format("%,d", 2000000)
outputs:
2,000,000

Once you've converted your String to a number, you can use
// format the number for the default locale
NumberFormat.getInstance().format(num)
or
// format the number for a particular locale
NumberFormat.getInstance(locale).format(num)

I've created my own formatting utility. Which is extremely fast at processing the formatting along with giving you many features :)
It supports:
Comma Formatting E.g. 1234567 becomes 1,234,567.
Prefixing with "Thousand(K),Million(M),Billion(B),Trillion(T)".
Precision of 0 through 15.
Precision re-sizing (Means if you want 6 digit precision, but only have 3 available digits it forces it to 3).
Prefix lowering (Means if the prefix you choose is too large it lowers it to a more suitable prefix).
The code can be found here. You call it like this:
public static void main(String[])
{
int settings = ValueFormat.COMMAS | ValueFormat.PRECISION(2) | ValueFormat.MILLIONS;
String formatted = ValueFormat.format(1234567, settings);
}
I should also point out this doesn't handle decimal support, but is very useful for integer values. The above example would show "1.23M" as the output. I could probably add decimal support maybe, but didn't see too much use for it since then I might as well merge this into a BigInteger type of class that handles compressed char[] arrays for math computations.

you can also use the below solution
public static String getRoundOffValue(double value){
DecimalFormat df = new DecimalFormat("##,##,##,##,##,##,##0.00");
return df.format(value);
}

public void convert(int s)
{
System.out.println(NumberFormat.getNumberInstance(Locale.US).format(s));
}
public static void main(String args[])
{
LocalEx n=new LocalEx();
n.convert(10000);
}

You can do the entire conversion in one line, using the following code:
String number = "1000500000.574";
String convertedString = new DecimalFormat("#,###.##").format(Double.parseDouble(number));
The last two # signs in the DecimalFormat constructor can also be 0s. Either way works.

Here is the simplest way to get there:
String number = "10987655.876";
double result = Double.parseDouble(number);
System.out.println(String.format("%,.2f",result));
output:
10,987,655.88

The first answer works very well, but for ZERO / 0 it will format as .00
Hence the format #,##0.00 is working well for me.
Always test different numbers such as 0 / 100 / 2334.30 and negative numbers before deploying to production system.

According to chartGPT
Using DecimalFormat:
DecimalFormat df = new DecimalFormat("#,###.00");
String formattedNumber = df.format(yourNumber);
Using NumberFormat:
NumberFormat nf = NumberFormat.getNumberInstance();
nf.setGroupingUsed(true);
String formattedNumber = nf.format(yourNumber);
Using String.format():
String formattedNumber = String.format("%,.2f", yourNumber);
Note: In all the above examples, "yourNumber" is the double value that you want to format with a comma. The ".2f" in the format string indicates that the decimal places should be rounded to 2 decimal places. You can adjust this value as needed.