[SOLVED] Why does this variable expansion with evals produce different numbers of backslashes?

Issue

[user@localhost ~]$ declare -p test
declare -- test="eval \\\\\\\\; eval \\\\\\\\;"
[user@localhost ~]$ $test
-bash: \\: command not found
-bash: \: command not found

It seems like the second set of backslashes is being parsed twice as much as the first set. This happens for any number of backslashes I add.

Why?

Solution

It's easier to see the problem with echo:

$ declare -- test="echo \\\\\\\\; echo \\\\\\\\;"
$ $test
\\\\; echo \\\\;

As you can see it does not run two echo commands, it runs a single echo and passes the rest of the string to it. This happens because $test does not cause shell code to evaluate. It will instead do word splitting into a so-called simple command and invoke the result.

To get the effect you expected, evaluate it as a shell string:

$ eval "$test"
-bash: \: command not found
-bash: \: command not found

To reproduce the result without the variable, where one eval invokes the second one with one after interpreting the backslashes:

$ 'eval' '\\\\;' 'eval' '\\\\;'
-bash: \\: command not found
-bash: \: command not found

$ eval '\\\\; eval \\\\;'  # Equivalent but more canonical
-bash: \\: command not found
-bash: \: command not found

Answered By - that other guy

Answer Checked By - Senaida (WPSolving Volunteer)