Why does $(RANDOM) work differently from $(ls -la)?

Question

for command substitution in bash, we do this

hello=$(ls -la)  
echo "$hello"

but when i do the same for below command,it fails,it says RANDOM not found

hell=$(RANDOM)
echo "$hell"

I can get over it ,by using arithematic expansion..

hell=$((RANDOM))
echo "$hell"

Any idea why below doesn't work...as per my understanding, RANDOM is also a command and i should have gotten number echoed

hell=$(RANDOM)
echo "$hell"

Answer 1

RANDOM isn't a command, it's a variable. Just one set by the shell and a bit of a special one:

RANDOM Each time this parameter^(*) is referenced, it expands to a random integer between 0 and 32767. Assigning a value to this variable seeds the random number generator. If RANDOM is unset, it loses its special properties, even if it is subsequently reset.

Something like echo "$RANDOM" is exactly how you'd use any variable, like the more "normal", HOME and BASH_VERSION. Of course ${RANDOM}, with braces and not regular parentheses works too.

Being able to use it in an arithmetic expansion without the $ is also a feature of the arithmetic context, and works for any variable:

Within an expression, shell variables may also be referenced by name without using the parameter expansion syntax.

(e.g. x=4; echo "$((x*x))" prints 16)

_{(* You can ignore the variable vs. parameter distinction here.)}