Prevent command substitution from running when declaring a variable

Question

So i recently noticed that, when declaring a variable using a command substitution, let's say like so:

var=$(echo "this is running" > test; cat test)

Then it will run (the file test will be created) , even though i didn't yet called it (technically), which i would like so:

var=$(echo "this is running" > test; cat test)
echo "$var" # is where i would "normally" call the variable

How could I prevent the command substitution from actually running when declaring it in a variable, so it only actually run when I call the said variable?

PS: Well aware this is kind of a bad example, but it serve well to demonstrate what i mean, although with "useless use of cat" and "useless use of echo"...

What other approach (beside using a function) could i use? just curious :) (don't mind using a function but would usually do that only if it's for a decently sized logic/command) @Quasímodo — Nordine Lotfi, Mar 23 '21 at 12:51
Yeah, i did thought of that (thought didn't know this was the normal behavior, which is why i asked initially) @FelixJN — Nordine Lotfi, Mar 23 '21 at 12:51
Similar: [Output to variable with an indirect command (eval)](//unix.stackexchange.com/q/205533) — Stéphane Chazelas, Mar 23 '21 at 12:53
What you are dealing with is a string that represents a piece of shell code that you want to run _later_. This is the typical use for functions. — Kusalananda, Mar 23 '21 at 12:53
Also: [How to implement "generators" like $RANDOM?](//unix.stackexchange.com/q/254962) — Stéphane Chazelas, Mar 23 '21 at 12:54
Mainly in case where i use said command substitution more than once AND in cases where said command substitution is too long to be repeated/typed more than once (in which case, i guess the logical thing would be a function...) @Quasímodo — Nordine Lotfi, Mar 23 '21 at 12:54
@NordineLotfi I think you are just mistaken: A (bash) variable is always a string. There is no `var=$something` that would dynamically return the current value of `$something` - `$var` will be set to the current value of `$something` and that's it. Anything dynamic must be achieved otherwise (e.g. function). — FelixJN, Mar 23 '21 at 12:55
See also: [What shells support "value substitution" like MirBSD Korn Shell?](//unix.stackexchange.com/a/611364) — Stéphane Chazelas, Mar 23 '21 at 12:56
@FelixJN, bash variables can be string or array or associative array or namerefs. There are special variables like `$SECONDS`, `$RANDOM`, `$PWD`... whose contents is automatically generated. You can also set the `integer` attribute on a variable which causes its contents to be interpreted as arithmetic expression (though in the case of `bash`, that's done upon assignment only *after* the attribute has been set; still, see `a='b[$(uname>&2)]' bash -c 'var=$a; typeset -i var; var=var+1'` for instance). — Stéphane Chazelas, Mar 23 '21 at 14:39
Your wording seems strange to me. You use the word 'call' with a variable, but you don't call a variable, you call a function. You assign a value to the variable in the first line which runs the code and you use the value assigned to that variable later. — Jason Goemaat, Mar 23 '21 at 22:39
Yeah, I'm aware i may not use the right word for the right term, but at least (seeing the other answer/replies) i can see my thought/right meaning were conveyed, so no harm done i believe :) @JasonGoemaat I appreciate that you corrected me though — Nordine Lotfi, Mar 23 '21 at 22:42

Stéphane Chazelas · Accepted Answer · 2021-08-10T10:39:59.350

Sound like you want a variable whose contents is dynamically generated.

bash does not have support for ksh93's disciplines, or zsh's dynamic named directory or mksh's value substitution which would make it easier, but you could use this kind of hack, using namerefs:

var_generator() { date --iso-8601=ns; }
var_history=()
typeset -n var='var_history[
  ${##${var_history[${#var_history[@]}]=$(var_generator)}},${#var_history[@]}-1
]'

Here with var defined as a reference to an element of the $var_history array, using the fact that array indices are evaluated dynamically and allow running arbitrary code (here used to run the var_generator function and assign its output to a new element of the array).

Then:

bash-5.1$ echo "$var"
2021-03-23T13:36:43,243211696+00:00
bash-5.1$ echo "$var"
2021-03-23T13:36:45,517726619+00:00

That sounds a bit too convoluted though where you could just use $(var_generator) here. One advantage though is that you can still do things like ${var#pattern} while bash (contrary to zsh) won't let you do ${$(cmd)#pattern}.

In your example, can i just replace the `var_generator` by the actual command, or I'm i obliged to put the command i want to use in a command substitution inside a function? (like you did with `var_generator`) — Nordine Lotfi, Mar 23 '21 at 14:27
@NordineLotfi, yes, you can put the command directly in there without using an intermediary function. Beware of quoting though. — Stéphane Chazelas, Mar 23 '21 at 14:32

score 1 · Answer 2 · answered Aug 10 '21 at 11:08

It runs because you did call it. Shell expansions happen right there, unlike e.g. variables in Make, which usually are lazily evaluated. If you did var='$(...)', the command substitution wouldn't run, but it would be difficult to run it later, either. This is similar to how something foo=$bar copies the value of bar at that moment, instead of looking at it later when foo is eventually used.

The simple alternative is to use a function instead. E.g. this should print two dates that are around 5 seconds apart:

getdate() {
    date +"%F %T"
}
a=$(getdate)
sleep 5
b=$(getdate)
echo "$a $b"

The downside of using a function is that the string operations you can use with parameter expansion aren't available. You can't do ${(getdate)% *} to get just the first part of the output, you'll have to do foo=$(getdate); foo=${foo% *};

Prevent command substitution from running when declaring a variable

2 Answers2

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