using a Bash variable in place of a file as input for an executable

Question

I have an executable that is used in a way such as the following:

executable -v -i inputFile.txt -o outputFile.eps

In order to be more efficient, I want to use a Bash variable in place of the input file. So, I want to do something like the following:

executable -v -i ["${inputData}"] -o outputFile.eps

Here, the square brackets represent some clever code.

Do you know of some trick that would allow me to pipe information into the described executable in this way?

Many thanks for your assistance

Answer 1

You can use the following construct:

<(command) 

So, to have bash create a FIFO with the command as the output for you, instead of your attempted -i ["${inputData}"], you would do:

-i <(echo "$inputData") 

Therefore, here is your final total command:

executable -v -i <(echo "$inputData") -o outputFile.eps 

Answer 2

Echo is not safe to use for arbitrary input.

To correctly handle pathological cases like inputdata='\ntest' or inputdata='-e', you need

executable -v -i <(cat <<< "$inputData")

In zsh, the cat is not necessary

---

Edit: even this adds a trailing newline. To output the exact variable contents byte-by-byte, you need

executable -v -i <(printf "%s" "$inputData")

Answer 3

Note: zsh only:

To get a filename containing the contents of ${variable}, use:

<(<<<${variable})

Note:

• <<<${variable} redirects STDIN to come from ${variable}
• <<<${variable} is equivalent to (but faster than) cat <<<${variable}

So for the OP's case:

executable -v -i <(<<<${inputData}) -o outputFile.eps