For example:
#!/bin/bash
sss='ls -l'
$sss
ttt='a=100'
$ttt
The output of ls is correct, however, the assignment statement will output an error message:
line 5: a=100: command not found
Why the difference?
If assignment is not command, what is it? I mean what is the difference between explicit a=100 and a=100 expanded from variable, I mean, the bash sees the same thing a=100, right? Why they got different interpretation?
That's because the output from variable expansion is run as a command, precisely get replaced in the command line as if you have inserted the content literally.
Here, you have ttt='a=100', so when you do $ttt next, it will be simple expanded as a=100 and as this would be the command to run being the only parameter present. And the error is due to the obvious fact that this is not a valid command.
You can tack the expansion with some actual-valid command to get the expansion as that command's argument (e.g. echo $ttt), as you can imagine.
If you ever need to do assignments like that, leverage declare:
$ ttt='a=100'
$ declare "$ttt"
$ echo "$a"
100
If you want to execute code, use a function, not a parameter expansion.
$ sss () { ls -l; }
$ sss
foo.txt bar.txt
$ ttt () { a=100; }
$ ttt
$ printf '%d\n' "$a"
100
Because a=100 is not a command, and you can't evaluate assignments like that.
In order to properly evaluate the assignment try:
eval $ttt
a=100 and a=100 expanded from variable, I mean, the bash sees the same thing a=100, right? Why it got different interpretation?
command for shell is one of alias, builtin, function, or an executable file either specified through a path or looked up through the list of directories specified by $PATH. In your case, ls was looked up and found in one of the directories of $PATH, however, a=100 wasn't found. Hence, 'command not found` error.
eval to a newbie?