Why does $(RANDOM) work differently from $(ls -la)? for command substitution in bash, we do this hello=$(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"

`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.)