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Why do we use hex representation as default for the output of a hash function's result?

For example, the SHA-256 hash function: the output of SHA-256 in hex representation uses 64 characters, while using Base64 on the raw output produces 44 characters.

Demo:

<?php
$password = "password";
$sha256 = hash('sha256',$password);
echo 'sha256('.strlen($sha256).'): '.$sha256.'<br />';

$sha256Base64 = base64_encode(hash('sha256',$password,true));
echo 'sha256('.strlen($sha256Base64).'): '.$sha256Base64.'<br />';

Output:

sha256(64): 5e884898da28047151d0e56f8dc6292773603d0d6aabbdd62a11ef721d1542d8
sha256(44): XohImNooBHFR0OVvjcYpJ3NgPQ1qq73WKhHvch0VQtg=
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  • $\begingroup$ My guess: because hex is more common for people and usually easier to understand and parse than base64. $\endgroup$
    – SEJPM
    May 1, 2016 at 19:25
  • $\begingroup$ This is more of an implementation question, and not really cryptography I think - but many programming languages understand hex, while base64 usually is a higher level function. Note that some cryptographic protocols use other bases since they give shorter representations (see e.g. base 58 in Bitcoin and Monero) $\endgroup$
    – aegbert
    May 1, 2016 at 19:40
  • $\begingroup$ In addition to all mentioned reasons, it's easy to see how many bytes a given hex string represents. $\endgroup$ May 4, 2016 at 2:37
  • $\begingroup$ @Dodekeract Yes, I also noticed that a sha256 base64 representation of a string does not always produce 44 characters, sometimes more (example, 58 characaters) $\endgroup$ May 4, 2016 at 11:25
  • $\begingroup$ @maarten seriously? do you have an example for this? just by the workings of how b64 works with using 6 bit on each character, you have a little over 2 by deviding 256 by 6, and since you cant have half a char you obviously go to 43, and then the equal sign at the end is just padding because b64 encodes 3 byte to 4 chars and therefore the byte count (32) needs to be divisible by 3 which means 1 passing byte gets added. this means it wouldnt be possible to have a Sha256 with more than 44 chars in b64 $\endgroup$
    – My1
    Sep 21, 2017 at 10:53

3 Answers 3

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Hexadecimal is traditional -- by this, I mean that there first were command-line tools that used hexadecimal for output, then other people using the hash functions found it fit to stick to hexadecimal, if only to be able to compare their values with the output of the aforementioned tools. That's how traditions get established: a more-or-less random choice at the start, then the need for interoperability and backward compatibility kicks in.

In the case of hexadecimal in cryptographic algorithms, one can probably trace it to the use of C language for reference implementations. Most algorithms are described with a specification (mathematical description, usually typeset in LaTeX), and a reference implementation that produces basic test vectors. For better or worse, the reference implementation is usually in C (or sometimes C++). In C, there is no standard facility for Base64 encoding (some programming platforms offer that, or external libraries, but it is not standard); but hexadecimal is easily obtained with a simple printf() with a "%08x" format string. As a very classic example, consider the MD5 specification (RFC 1321), which contains a reference implementation that does hexadecimal output.

The tradition is well entrenched; for the SHA-3 competition, NIST actually asked for reference implementations in C, and known-answer tests with a fully-specified text format that was hexadecimal throughout.

It must also be said that hexadecimal is convenient for debugging: the human developer can easily observe hexadecimal output and map these to individual bits, by doing the simple conversion in his head. Base64 is not as simple, because it entails 64 glyphs instead of 16, including some which are prone to induce visual confusion (1 vs I vs l, 0 vs O...). Also, many algorithm internally use 32-bit or 64-bit words, that map well to CPU registers; 32 and 64 are multiples of 4 but not of 6, so Base64 encoding again implies some non-trivial splitting.

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  • $\begingroup$ IBM S/360 and its assembler BAL 'standardized' 8-bit bytes in hex (including BCD and floating-point using 4-bit fields) for much of the computer industry almost a decade before C. And I believe (but can't find a good source) Lucifer started as an option on S/360. OTOH C supports octal about equally to hex -- and did you know that if you give an IPv4 address string like 010.020.030.040 to most Unix software (starting with BSD) it's octal and means 8.16.24.32? $\endgroup$ May 6, 2016 at 8:24
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In short: Hexadecimal is virtually a gold standard for radix 16 encoding. Base64 isn't standard at all.

Hex (quoting):-

the letters A–F or a–f represent the values 10–15, while the numerals 0–9 are used to represent their usual values.

And each character represents a nibble. So exactly two characters per byte.

Now consider Base64. There may or may not be padding. The 62nd and 63rd characters can vary according to protocol. Sometimes there's even a cyclic redundancy check automatically included. Let me just list part of the Base64 Wiki page contents:-

3 Examples

    3.1 Output padding
    3.2 Decoding Base64 with padding
    3.3 Decoding Base64 without padding

4 Implementations and history

    4.1 Variants summary table
    4.2 Privacy-enhanced mail
    4.3 MIME
    4.4 UTF-7
    4.5 OpenPGP
    4.6 RFC 3548
    4.7 RFC 4648
    4.8 The URL applications
    4.9 HTML
    4.10 Other applications
    4.11 Radix-64 applications not compatible with Base64

It's a protocol dependent mess. And notice §4.11! So it's just simpler and less prone to implementations/interpretations and variations/errors.

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Because in JS, numbers are always a representation of a 64-bit floating point number (the bit, mantissa and exponent), and on the SHA-256 hash function, hex, which is also equivalent to 16, always generates a 64-bit encoded representation.

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  • $\begingroup$ Cryptography is independent of any programming language if you mean JavaScript by JS. Also, you're confusing 64-bit and 64-character. $\endgroup$
    – DannyNiu
    Jul 17, 2021 at 10:05

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