Random Java examples - comments, issues, pull requests welcome.
int a, b;
for (a=b=(1<<11)-1; a<1<<(11<<1); a<<=1,b>>=1)
System.out.println(Integer.toString(
(1<<11)|(a^b)&((1<<11)-1),1<<1).replace("1"," "));The result of running that program may not be obvious.
It uses two int variables, a single loop, a bunch of bitwise operators
and only digit 1 in literals.
int m = 5;
for (int j = -m; j <= m; j++) {
for (int i = 0; i-m <= Math.abs(j); i++)
System.out.print(m-i > Math.abs(j) ? " " : "*");
System.out.println();
}Sometimes absolute values in loops can be useful.
LoopArithmetic.java (download)
int m = 5;
for (int j = 0; j <= 2*m; j++) {
for (int i = 0; i <= 2*m; i++)
System.out.print((i-j)*(2*m-j-i) < 0 ? " " : "*");
System.out.println();
}Using arithmetic operations to simplify loops at the expense of readability.
StringsAndMirrors.java (download)
static String mirror(String s) {
return mirror(s, "");
}
static String mirror(String s, String m) {
return s.length() == 0 ? m :
mirror(s.substring(1), s.substring(0,1) + m);
}Using method overloading to simplify recursion.
Not as nice as StringsAndMirrors.js (download):
function mirror(s) {
return s.split('').reverse().join('');
}but still pretty good for Java.
(Note: I posted a question about this example on Stack Exchange Code Review: Compact ways to filter even numbers in Java int array - please post an answer if you can improve the code below)
static int[] even(int[] t) {
int i, j;
for (i=j=0; i < t.length; i++)
if (t[i] % 2 == 0) j++;
int[] r = new int[j];
for (i=j=0; i < t.length; i++)
if (t[i] % 2 == 0) r[j++] = t[i];
return r;
}Returns an array with only even numbers. It's very hard to write in a less verbose way in Java because of poor support for higher order functions and no built-in filter method for arrays. For a comparison, here is the same in JS:
function even(t) {
return t.filter(function(x){ return x % 2 == 0; });
}It's slightly shorter with ES6 syntax:
function even(t) {
return t.filter(x => x % 2 == 0);
}It's even better in Scheme:
(define (even t)
(filter even? t))Not bad for a language that's 40 years old.
static int[] eosort(int[] t) {
int i, ei, oi, l = t.length;
for (i=ei=0; i < l; i++)
if (t[i] % 2 == 0) ei++;
int[] r = new int[l],
e = new int[ei],
o = new int[l-ei];
for (i=ei=oi=0; i < l; i++) {
if (t[i] % 2 == 0) e[ei++] = t[i];
else o[oi++] = t[i];
}
java.util.Arrays.sort(e);
java.util.Arrays.sort(o);
for(i = 0, oi=o.length; i < oi/2; i++) {
int s = o[i];
o[i] = o[oi-i-1];
o[oi-i-1] = s;
}
for (i=ei=oi=0; i < l; i++)
r[i] = t[i] % 2 == 0 ? e[ei++] : o[oi++];
return r;
}This is a pretty compact way to do a two-way sorting of an array (even elements ascending and odd elements descending at the same time, preserving even/odd positions).
But what is written in Java in 19 lines (not including empty lines and the return statement) can be written in JS in just 3 lines:
function eosort(t) {
var b = [t.filter(x => x % 2 == 0).sort(),
t.filter(x => x % 2).sort().reverse()],
i = [0,0], r = t.map(x => b[x%2][i[x%2]++]);
return r;
}Here is also an pre-ES6 version of the above code that should conform to the original ES1 version of the language from 1997 (please post an issue if that is not the case). The only difference is more cluttered syntax but there are still exactly the same expressions:
function eosort(t) {
var b = [t.filter(function(x){ return x % 2 == 0; }).sort(),
t.filter(function(x){ return x % 2; }).sort().reverse()],
i = [0,0], r = t.map(function(x){ return b[x%2][i[x%2]++]; });
return r;
}It seems impossible to implement it in Java in a way even remotely as compact as in JS, which seems strange for a language as mature and popular as Java. If anyone can improve the Java code to make it more compact, please do a pull request because I want that comparison to be fair.
That example was not chosen specifically to favour JS. In fact it is a school example straight from a Java class.
Latest version in a ZIP file:
https://github.com/rsp/ppj/archive/master.zip
For any bugs please post an issue.
Rafał Pocztarski - https://github.com/rsp
MIT License (Expat). See LICENSE.md for details.