Java：将数字和字符串转换为十六进制会返回不同的结果。 为什么？

Question

Trying to convert numbers to hex got me stuck, take a look:

Number: 32

Hex expected (ASCII): 20

Result from number that came as string:

System.out.println(String.format("%02x", new BigInteger(1, "32".getBytes(StandardCharsets.US_ASCII))));

Gives me 33 32 as result (which I understood that happens because it parses 3 (=33) and 2 (=32) )

Result converting as number:

System.out.println(Integer.toHexString(32));

Gives me 20 (correct)

I would like to better understand this situation, can someone explain what differs from each other? (please don't say "because it calls different methods... be friendly)

Also, the first approach lets me set the Charset and the second one doesn't. Why?

Answer 1

return of getBytes() method from a String, separates every byte of that String which in your case would be '3' and '2'.

printing them using String.format("%02x", ..... , shows hexadecimal number of their ASCII codes.

in the other hand, Integer.toHexString(32) gets data from toHexString method and this method does not separate every single byte and gets data in hexadecimal format.