Profile and optimizeyour Java code

Gabriel Laden

CS 146 – Dr. Sin-Min Lee

Spring 2004

Introduction

• O(n) notation is one thing

• Performance of your coded algorithm is another thing!!!

• This demo will use Homework #3 “ICBaseCycle” problem

Where to start testing?

• To find run-time between two sequential lines in your code insert the following:

System.currentTimeMillis();

The 80/20 rule

• Your code spends 80% of its time in 20% of the code, or so they say…

• How to find this hotspot in your code without just guessing :

java –Xprof class [args]

java –Xprof class [args]

• -Xprof is a built-in profiler for your code

• It tells you what % time is spent in running and other Java administrative tasks

• Caution: eliminate your user I/O and other things like that which are not constant and will throw off your testing results

Other interesting java –X options

• -Xloggc:<file> log GC status to a file with time stamps

Note: see use later in presentation

• -Xmx<size> set maximum Java heap size

Note: 64M is default, good use is for problems of moderately larger size, (go buy more memory at Fry’s)

• -Xss<size> set java thread stack size

Note: good use in recursive problems

javap - The Java Class File Disassembler

javap [options] class• The javap command disassembles a class file. Its output

depends on the options used. If no options are used, javap prints out the package, protected, and public fields and methods of the classes passed to it.

• -c option prints out disassembled code, i.e., the instructions that comprise the Java bytecodes

Most Important Step!!!

• Find your inner-most loop that is in method with largest run-time %

• Re-write this loop to be most efficient

• This is where O(n) notation theory can be applied to your algorithm

Java handles your garbage

• So don’t forget to analyze this!!!• java -Xloggc:<file> class

• The cost of gc also implies the cost of the allocation

• The number of lines written to filegc will be evidence of the garbage collector activity

• Look in your code for unneccessary allocations

• Re-order processing to keep heap size to a minimum

Try a few different things

• Is your sort algorithm really the best one?

• Is your data structure really the best one?

• Save your working program to a backup!!!

• Compare run-times with different sorts

• Do the least # operations to get your output

Summary

• v1 original code

• v2 inner loop re-write / algorithm design

• v3 minimize allocations

• v4 improved version

(…repeat (1-4) if necessary)

• v5 final version

// version 1private boolean sumElements(int[] permutation){

//generate sumation of terms 1+2+3+...+pint walk, sum;int index = 0;int[] partSums = new int[p * (p-1) + 1];

// i = length of the elements of sumfor(int i=1; i<p; i++){

//j = starting indexfor(int j=0; j<p; j++){

sum = 0;walk = j;

//k = loop for summation for(int k=0; k<i; k++){

if(walk == p){walk = 0;

}sum += permutation[walk++];

}partSums[index++] = sum;

} }// (con’t next slide…)

// version 1 (con’t)

//add last element sum of all elements

partSums[index] = n;

//qsort rewritten from QSortAlgorithm.java in j2sdk demo directory

QSort.qsort(partSums, 0, partSums.length-1);

//step through array to verify success

int consecutive = 1;

for(int i=1; i<partSums.length; i++){

if(partSums[i] == consecutive + 1){

consecutive++;

}

else if(partSums[i] > consecutive + 1){

return false;

}

}

return true;

}

// version 5private boolean sumElements(int[] permutation, boolean[] partIndexes){

//generate sumation of terms 1+2+3+...+pint value, walk;int pMinusOne = p - 1;Arrays.fill(partIndexes, false);for(int i=0; i<p; i++){

walk = i + 1;partIndexes[(value = permutation[i])] = true;for(int j=1; j<pMinusOne; j++, walk++){

partIndexes[(value += permutation[walk])] = true;}

}//skip over index 0, not used//skip over index 1 & 2, always truefor(int i=3; i<partIndexes.length; i++){

if(partIndexes[i] == false){return false;

}}return true;

}

C:\cs146_profiling>java hw3.v1.ICBaseCycle 8 57Here is a set of valid arrays1 8 11 3 18 10 2 41 12 2 6 3 7 22 41 5 15 7 16 2 8 31 8 2 3 21 4 12 61 6 17 12 2 11 5 31 2 10 19 4 7 9 5

# of solutions: 6Total compute-time: 110.328 seconds

C:\cs146_profiling>java hw3.v5.ICBaseCycle 8 57Here is a set of valid arrays1 8 11 3 18 10 2 41 12 2 6 3 7 22 41 5 15 7 16 2 8 31 8 2 3 21 4 12 61 6 17 12 2 11 5 31 2 10 19 4 7 9 5

# of solutions: 6Total compute-time: 10.078 seconds

Conclusion

• Easy to start learning how to test the performance on our homework problems

• Later you will be running large scale applications with multi-tier enterprise applications with networking and database, etc. it will be a lot harder to start learning how to optimize then!!!

References

• Effective Java - Joshua Bloch

• Java Platform Performance – Steve Wilson

• Core Java v.2 – Cay Horstmann