Java Format string to C#

Question

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.

Also, your format string is a bit off. The first number after '{' is a placeholder, starting with 0. After the ':' you place the formatting. — Tim
– Tim, Commented Sep 2, 2020 at 7:57
The error is telling you the format string is wrong. You have three '{}', so it should be '"{0:fromatstring} {1:formatstring} {2:format string}'. In your posted string, the second placeholder starts with 7, when you only have 3 total placeholders. — Tim
– Tim, Commented Sep 2, 2020 at 7:58
See learn.microsoft.com/en-us/dotnet/api/… and learn.microsoft.com/en-us/dotnet/standard/base-types/… for more info. — Tim
– Tim, Commented Sep 2, 2020 at 8:01
For example, (not that familiar with Java), the first placeholder could be something like {0:d}, and with String.Format would be String.Format("{0:d}", 6); — Tim
– Tim, Commented Sep 2, 2020 at 8:02
PLEASE. PLEASE. PLEASE. DO not close this. Perhaps show a Java minimal reproducible example with actual inputs and actual output so we can see what you are trying to achieve in C#? — mjwills
– mjwills, Commented Sep 2, 2020 at 8:12

Tim · Accepted Answer · 2020-09-02 09:03:33Z

2

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.

edited Sep 2, 2020 at 9:03

answered Sep 2, 2020 at 8:09

Tim

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4 Comments

John Smith Over a year ago

Hi Tim, %6d %7.1f %5.1f <-- how ? Because the way I understand it, is that those are replaced by other values in Java. Although I do not know what they actually mean.

Tim Over a year ago

I don't know much about Java, but I believe that's just Java's way of formatting strings. It'll take the 6 and format as a decimal, the 7.1 as a float, etc. The above code snippet will return "6 7.10 5.10" (because I didn't indicate the precision. What does the output in Java look like?

mjwills Over a year ago

@Tim It's obvious what values will be generated based on variable data types. is what the OP said the last time we asked to see the output.

Tim Over a year ago

@mjwills I'm not familiar enough with Java's string formatting to say for certain (its also past my bedtime so maybe my brain shut off). The double/double thing confuses me, though I've been able to figure out %6d works with the length of the integer passed in. That's why I edited my answer to see it's a very basic intro to String.Format in C#.

Chandra Prakash Ajmera · Accepted Answer · 2020-09-02 08:47:19Z

0

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);

answered Sep 2, 2020 at 8:47

Chandra Prakash Ajmera

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2 Comments

Chandra Prakash Ajmera Over a year ago

Welcome! and happy to help :)

Tim Over a year ago

This answer doesn't take into account the padding and precision in the Java formatting. May or may not be of importance to the OP.

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