convert a binary String containing 64bits to hexadecimal in java?

Question

I'm trying to convert this binary String that contains 64bits to Hexadecimal with this code:

String mm = "1000010111101000000100110101010000001111000010101011010000000101";
String v = new BigInteger(mm, 2).toString(16);
v=String.format("%64x", v);

but it gives me this exception:

Exception in thread "main" java.util.IllegalFormatConversionException: x != java.lang.String
    at java.util.Formatter$FormatSpecifier.failConversion(Unknown Source)
    at java.util.Formatter$FormatSpecifier.printInteger(Unknown Source)
    at java.util.Formatter$FormatSpecifier.print(Unknown Source)
    at java.util.Formatter.format(Unknown Source)
    at java.util.Formatter.format(Unknown Source)
    at java.lang.String.format(Unknown Source)
    at test.main(test.java:332)

what is getting wrong and why?

That looks like a binary string rather than a hex string. Why the 16 in the BigInteger constructor? Shouldn't it be 2? — John Coleman
– John Coleman, Commented Mar 10, 2017 at 1:27
@JohnColeman Sorry, you are right, I had edit it and forgot to change when I have post this question. Thank u. — Sivar
– Sivar, Commented Mar 10, 2017 at 1:29
Your String.format("%64x", v); uses a type character of x which says to treat the argument v as an unsigned int and format it in hexadecimal — but your v is a String already. — Stephen P
– Stephen P, Commented Mar 10, 2017 at 1:35

Ousmane D. · Accepted Answer · 2017-03-10 01:35:10Z

3

change this:

v=String.format("%64x", v);

to this:

v=String.format("%s", v);

Also you need this:

String v = new BigInteger(mm, 2).toString(16);

edited Mar 10, 2017 at 1:35

answered Mar 10, 2017 at 1:25

Ousmane D.

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2 Comments

Sivar Over a year ago

Thank you very much for the solution – it was great. :)

Scary Wombat Over a year ago

This just appends 64x onto the end of 85e813540f0ab405

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