How can this src vs. build tree timestamp comparison be faster?

Question

for n in `cd src; find . -name "*.java"; cd -`;
  do a=`echo $n | cut -d '.' -f2`;
  if [[ src/$a.java -nt build/$a.class ]];
    then echo src/$a.java;
  fi;
done

It lists all the java files in the src tree; then for each one, it removes the suffix ".java" (cut -d '.' -f2 because find . output is prefixed with .). It then uses -nt to test if the java file in the src tree is newer than the corresponding class file in the build tree; if it is newer, it is output. [javac can then use this to compile only the needed src files, instead of using ant or make]

The problem is that it is too slow, taking about 275ms. How to make it faster?

Or is there a faster way to do this in bash? I can't see how to do it with find, diff, rsync nor make (which doesn't seem to traverse trees automatically, and needs explicitly listed source files).

Answer 1

try this:

for f in src/*.java; do 
  file=`basename $f`; 
  if [[ src/$file -nt build/${file//.java/} ]]; then 
     echo src/$file; 
  fi; 
done