Make my pseudocode real

Question

I've got some Java pseudocode that uses whitespace instead of curly braces, and I want you to convert it.

I/O

Your program should take an input file along with a number designating how many spaces are used to indent a block. Here's an example:

$ convert.lang input.p 4
// Convert using 4 spaces as the block delimiter
$ convert.lang input.p 2
// Convert using 2 spaces as the block delimiter

It should then convert the result using the specified block delimiter and output the result to stdout.

The meat of the program

Blocks open with : and each line within the block is indented using the block delimiter, like Python code.

while(true):
 System.out.println("Test");

Each : is replaced with a {, and a } is appended to the end of the block.

while(true) {
 System.out.println("Test");
}

Examples

Input:

public class Test:
 public static void main(String[] args):
 System.out.println("Java is verbose...");

Output:

$ convert Test.pseudojava 4
public class Test {
 public static void main(String[] args) {
 System.out.println("Java is verbose...");
 }
}

Input:

main():
 printf("Hello World");

Output:

$ convert test.file 2
main() {
 printf("Hello World");
}

Input:

def generic_op(the_stack, func):
 # Generic op handling code
 b = the_stack.pop()
 if isinstance(b, list):
 if b:
 return
 top = b.pop(0)
 while b:
 top = func(top, b.pop(0))
 the_stack.push(top)
 else:
 a = the_stack.pop()
 return func(a, b)

Output:

$ convert code.py 4
def generic_op(the_stack, func){
 # Generic op handling code
 b = the_stack.pop()
 if isinstance(b, list) {
 if b {
 return
 }
 top = b.pop(0)
 while b {
 top = func(top, b.pop(0))
 }
 the_stack.push(top)
 }
 else {
 a = the_stack.pop()
 return func(a, b)
 }
}

Scoring

The code with the least amount of bytes wins!

• Martin Ender: Can we assume that the input contains no comments?
• jado: @MartinBüttner It may contain comments, but the comments won't contain ':'. Basically, yes.
• Peter Taylor: What about labels, which are the usual reason that a line would end in a colon in valid Java source?
• SuperJedi224: I've never seen a label anywhere but the beginning of a line.
• lirtosiast: I've just been reminded of how much I hate Java.

Answer 1

Perl, 41 bytes

#!perl -p0
1while s/( *).*\K:((
1円 .*)+)/ {2円
1円}/

Counting the shebang as two, input is taken from stdin. A command line argument need not be provided. Any valid nesting context can be determined (and matched) without knowing the indentation size.

---

Regex Break-Down

( *) # as many spaces as possible (1円)
 .* # as many non-newline characters as possible \
 (greediness ensures this will always match a full line)
 \K # keep all that (i.e. look-behind assertion)
 : # colon at eol (newline must be matched next)
 (
 (
 \n1円 # newline with at least one more space than the first match
 .* # non-newlines until eol
 )+ # as many of these lines as possible
 ) # grouping (2円)

---

Sample Usage

in1.dat

public class Test:
 public static void main(String[] args):
 System.out.println("Java is verbose...");

Output

$ perl py2java.pl < in1.dat
public class Test {
 public static void main(String[] args) {
 System.out.println("Java is verbose...");
 }
}

---

in2.dat

main():
 printf("Hello World");

Output

$ perl py2java.pl < in2.dat
main() {
 printf("Hello World");
}

---

in3.dat

def generic_op(the_stack, func):
 # Generic op handling code
 b = the_stack.pop()
 if isinstance(b, list):
 if b:
 return
 top = b.pop(0)
 while b:
 top = func(top, b.pop(0))
 the_stack.push(top)
 else:
 a = the_stack.pop()
 return func(a, b)

Output

$ perl py2java.pl < in3.dat
def generic_op(the_stack, func) {
 # Generic op handling code
 b = the_stack.pop()
 if isinstance(b, list) {
 if b {
 return
 }
 top = b.pop(0)
 while b {
 top = func(top, b.pop(0))
 }
 the_stack.push(top)
 }
 else {
 a = the_stack.pop()
 return func(a, b)
 }
}

• Not that Charles: i just wrote this in ruby

Answer 2

Python 3, (削除) 299 (削除ここまで) 265 bytes

import sys;s=int(sys.argv[2]);t="";b=0
for l in open(sys.argv[1]):
 h=l;g=0
 for c in l:
 if c!=" ":break
 g+=1
 if g/s<b:h=" "*g+"}\n"+h;b-=1
 if l.strip().endswith(":"):h=l.split(":")[0];h+=" {";b+=1
 t+=h+"\n"
b-=1
while b>-1:
 t+=" "*(b*s)+"}\n"b-=1
print(t)

Boom bam pow.

Algorithm used:

//global vars
string total //total modified program
int b //indent buffer
line thru lines: //iterate over every line
 string mline = "" //line to be added to the total
 //calculate the amount of spaces before the line (could be a lot easier)
 int spaces = 0 //total spaces
 c thru line: //go through every character in the line
 if c != " ": //if the current char isn't a space (meaning we went though all of them
 break //break out of iterating through chars
 spaces++ //increment the spaces because we've hit a space (hurr derr)
 if spaces/SPACE_SETTING < b: //if indent count of the current line is less than the indent buffer
 mline = "}\n" + line //add closing bracket to beginning of line
 b-- //decrement buffer
 if line.endswith(":"): //if the line ends with a `:`
 remove : from line
 mline += " {" //append {
 b++ //increment buffer
 total += mline //add modified line to total
print(total)

• TanMath: Can you include an explanation?
• jado: @TanMath Added the algorithm I used
• wb9688: Does it work with Python 2?
• jado: @wb9688 No idea, I only tested it with Python 3
• wb9688: I just tested, and it works with Python 2

Answer 3

Ruby, 70

x=$_+"
"
1while x.sub! /^(( *).*):
((2円 .*?
)*)/,'1円 {
3円2円}
'
$><<x

Adds a trailing newline. Does not need the indent block-size parameter.

Run this with -n0 (this is really 68+2). Thank you greatly to @primo for saving over a dozen bytes.

• primo: I think -p0 also works for ruby (-0 reads all input at once, -p stores stdin into $_, and auto-prints it at the end).
• Not that Charles: @primo Good point. In case it wasn't clear with my comment above, this isn't a port of your code, but my own work that does exactly what yours does (but with more bytes)
• primo: I understood correctly, just providing a tip to remove the (rather verbose) x=$<.readlines*''. While I'm doing that, sub! also has a two parameter overload (rather than one paramater + block) that accepts a replacement string, so you can use 1円, 2円, etc. instead of needing to concatenate it all.
• Not that Charles: @primo Thanks! I tried the two-parameter version before and abandoned it at some point last night.