# Why do we use hex output for hash functions?

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

For example the sha256 hash function: the output of sha256 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=
• My guess: because hex is more common for people and usually easier to understand and parse than base64. – SEJPM May 1 '16 at 19:25
• 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) – aegbert May 1 '16 at 19:40
• In addition to all mentioned reasons, it's easy to see how many bytes a given hex string represents. – Florian Wendelborn May 4 '16 at 2:37
• @Dodekeract Yes, I also noticed that a sha256 base64 representation of a string does not always produce 44 characters, sometimes more (example, 58 characaters) – swordsecurity May 4 '16 at 11:25
• @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 – My1 Sep 21 '17 at 10:53