如何使用sed删除Java项目的注释？

Question

I have a Java project, in which I have JavaDoc comments

/** ... */

other multi-line-comments

/* ... */

line comments

// ...

and my own "explanatory comments"

//* ...

When I release my code, I would like to have all line comments removed – not the other comments though. I though I would do it with sed, but so far I have not been successful. I am trying the following:

#!/bin/bash

while read -d $'\0' findfile ; do
  echo "${findfile}"
  mv "${findfile}" "${findfile}".veryold
  cat "${findfile}".veryold | sed -e 's|//[^\*"]*[^"]*||' -e 's/[ ^I]*$//' | grep -A1 . | grep -v '^--$' > "${findfile}"
  rm -f "${findfile}".veryold
done < <(find "${1}" -type f -print0)

What am doing wrong? Note that // in "..." should not be removed, since they might be part of a URL.

The crucial part is

-e 's|//[^\*"]*[^"]*||'

Answer 1

For a test file that looks like this:

/** This should stay */
/* And this
 * should stay
 * as well */
// This one should be removed
//* But this one should stay
code here // This part should go, but not the next line
"http://test.com"
code here //* This should stay

you can do it like this:

$ sed '\#^//[^*]#d;s#//[^*"][^"]*$##' test.java
/** This should stay */
/* And this
 * should stay
 * as well */
//* But this one should stay
code here
"http://test.com"
code here //* This should stay

The first expression, \\#^//[^*]#d , deletes all lines that start with // (but not //* ). This is to avoid getting empty lines in the output when the whole line is removed.

The second expression, s#//[^*"][^"]*$## , matches from // on (but not //* or //" ), until the end of the line, unless there is a " between the // and the end of the line.

Your expression s|//[^\\*"]*[^"]*|| did almost the same, except:

• No need to escape characters in bracket expressions; [\\*] matches both \\ and * .
• You want the first bracket expression to match just once, not zero or more times; your expression does match //* .
• You don't anchor at the end of the line, so this expressions matches // everywhere and removes them, even if followed by " or * .

For the peculiar case of a zero-length comment at the end of the line,

code here //

where you'd want to remove the // , you'd have to add a third substitution s#//$/# because the existing one expects at least one character after the // .

Notice how this is not super clean in that it might leave useless spaces at the end of lines, but that can easily be remedied by s/[[:space:]]*$/$/ .