Why is the following not working (no output):
fileRe="-regex '.*\.h'"
find "." -type f $fileRe
but the following works (and outputs results)
find "." -type f -regex '.*\.h'
Working with: /c/Program Files/Git/bin/bash
this is not intuitive?
1 Answer
When you type find "." -type f -regex '.*\.h' and press Enter,
the shell interprets the single quotes as the enclosing markers of a fixed string, and uses the enclosed value as the argument to pass to the find command.
In contrast with that, when you set fileRe="-regex '.*\.h'",
the value within the double-quotes is stored as-is (after variable expansion), and the embedded single-quotes are taken literally.
And then, when you pass $fileRe to find,
the value is taken as it was stored,
it is not reinterpreted,
so the literal single-quotes in the value remain literal single-quotes.
To achieve what you want, you can use a Bash array instead:
fileRe=(-regex '.*\.h')
find . -type f "${fileRe[@]}"
The difference in this version from yours is that the arguments passed to find are -regex and .*\.h. The single-quotes are not part of the value, because they were used when creating the value of the array, and they were not embedded in a double-quoted expression.
This behavior is the same in any Bash shell, regardless of the operating system.
fileRe=( -regex '.*\.h' )andfind . -type f "${fileRe[@]}".fileRe=( -regex "'$1'")I want to quote the expanded arguement $1 ?? This does not workfileRe=( -regex "$1" )and expand the array as"${fileRe[@]}"(with the double quotes).