bash中的\\ $$有什么用？

Question

I found this as a suggestion of how to store the output of "eval" into a variable called line. So, what's the use of \\$$ ?

command = "some command"
line = $(eval \$$command)

Answer 1

The \\$ prevents the shell from trying to treat the $ as the beginning of a parameter expansion. However, the code as a whole doesn't do anything useful. After fixing the whitespace issues and adding a real command to the example, your code looks like

command="ls -l"
line=$(eval \$$command)

command is simply a string ls -l . To evaluate the next line, the shell first evaluates the command substitution. The first step is to expand the parameter command , yielding line=$(eval \\$ls -l) . Quote removal gets rid of the backslash, so eval receives the arguments $ls and -l . Since ls presumably is not a variable, $ls is expanded to the empty string, and eval is left simply with -l to execute. There being no such command, you get an error.

You might think, then, that the correct form is simply

line=$(eval $command)

or slightly better

line=$(eval "$command")

That will work for simple cases, but not in general. This has been hashed over many times in many questions; see Bash FAQ 50, "I'm trying to put a command in a variable, but the complex cases always fail!" for the details.

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To answer the literal question, though, \\$$ is useful for outputing the string $$ , instead of expanding it to the current process ID:

# The exact output will vary
$ echo $$
86542

# Literal quotes
$ echo \$\$
$$

# Escaping either quote is sufficient
$ echo \$$ $\$
$$ $$