How can I check the Base64 value for an integer?

Question

I would like to check the Base64 value for an integer. There is a base64 linux command but I don't understand how I can apply it on integers.

I have tried with base64 10 but then I get the error message base64: 10: No such file or directory

I think that the problem can be that Base64 is used for Binary to Textual conversion, so there is no point to pass a Textual argument to it. So my argument 10 is read as an textual string and not as a binary number. Is there any way I can turn a texttual argument to a binary?

How can I use the base64 command to get the Base64 value for an integer?

Answer 1

Convert the number into hex than use echo to print the according byte sequence and pipe that into base64. So to encode the integer 10 with base64, you can use:

echo -en '\xA' | base64

To explain the result. The byte 10 has the following binary representation:

00001010

What base64 does is, it chunks those into groups of 6 bits. So with padding we get the following two 6lets:

000010 100000

Which in decimal are 2 and 32, which correspond to the letters C and g.

Answer 2

The command below converts a decimal integer to radix 64. It does not use the Base64 encoding scheme, but it uses the characters used by the URL-safe variant of the Base64 encoding scheme as digits.

awk 'NR==FNR{a[NR-1]=$0;next}{if($0==0){print"A";next}o="";for(n=$0;n!=0;n=int(n/64))o=a[n%64]o;print o}' <(printf %s\\n {A..Z} {a..z} {0..9} - _) -

The command below does the same in Bash. It also works with Zsh if you replace l%64 with l%64+1.

unset a;a=({A..Z} {a..z} {0..9} - _);while IFS= read l;do if ((l==0));then echo A;continue;fi;o=;for ((;l!=0;l/=64));do o=${a[l%64]}$o;done;printf %s\\n "$o";done

This converts radix 64 to decimal:

awk 'NR==FNR{a[$0]=NR-1;next}{o=0;for(i=NF;i>=1;i--)o+=a[$i]*64^(NF-i);print o}' FS= <(printf %s\\n {A..Z} {a..z} {0..9} - _) -

This does the same in Bash:

unset a;i=0;declare -A a;for x in {A..Z} {a..z} {0..9} - _;do a[$x]=$((i++));done;while IFS= read l;do o=0;for((i=${#l};i>=1;i--));do let o+=${a[${l:i-1:1}]}*64**(${#l}-i);done;echo "$o";done

The awk commands result in loss of precision with 2**53+1 and larger numbers, but to avoid it you can add -M (--bignum) in gawk 4.1 and later.

The Bash commands result in integer overflow with 2**63 and larger numbers.