What is the meaning of this pattern arg="${arg//\\/\\\\}"

Question

Please explain the use of pattern

arg="${arg//\\/\\\\}"

Does it mean exact match?

Answer 1

${var//pattern/replacement} is a parameter expansion operator of the ksh93 shell.

It expands to the value of $var where every occurrence of strings matching the pattern have been replaced with replacement.

That operator has been copied by a few shells including zsh, bash and mksh. It is otherwise not a standard sh parameter expansion operator.

The pattern is a glob pattern of the shell for instance where ? matches a single character, * matches any sequence of characters, [abc] match any one of a, b or c characters (and various shells have additional extensions over those).

If a special glob character is quoted like with '*', \*, "*"¹, then it is taken literally. ${var//\?/replacement} for instance replaces question marks instead of every single character.

A backslash must also be quoted like with \\ or '\' or "\\" to represent itself so ${var//\\/\\\\} means expand to the value of $var where each occurrence of backslash has been replaced with two backslashes.

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^{¹ Using \ as opposed to other quoting operators is generally preferable though for the cases where the whole ${...} is within double quotes. If the patterns and/or replacements are derived from expansions, and those expansions are not quoted (like in ${var//$pattern/$replacement} as opposed to ${var//"$pattern"/"$replacement"}, then backslashes must be doubled in ksh93 and bash, but must not in mksh or zsh (at least in their current versions)}