EMU8086 help

Modify the program so that it prints the decompressed data.

The compressed data is 1, 4, 3, 9, 5, 3, 2, 7, 0.  The program currently just prints all these values except the terminating zero.  When properly decompressed, the output should be:

 4 9 9 9 3 3 3 3 3 7 7

Modify the program to print out the decompressed data.  To do this you need to realize that the data comes in pairs like 5, 3.  This first value in the pair is the number of times the second value is to be repeated.  So set up a nested, inner loop to execute this number of times, and print the second value in the pair inside this inner loop.


The dumbfuck didn't give much of an explination on how to do it.

It would help if you posted the program you're supposed to modify.

That's the problem. The guy didn't leave me the program to modify. Asshole! Anyways, I'll provide the picture by my photobucket:

http://i166.photobucket.com/albums/u84/roy9076/rle_soln.gif

>>3
That essentially contains the full solution. You only need to add code for the branches.

I see you like lolis.

Anyone can help me about the coding?

>>5
/prog/ is not your personal army.

                       //`'''```,
             o        // LISP   `.,
       ,....OOo.   .c;.',,,.'``.,,.`

>>3
I see you're a weeaboo faggot.

Anyone can help me about the coding?
$ perl -le'print join" ",map{($,$v)=(shift,shift);map$v,1..$c}@ARGV' 1 4 3 9 5 3 2 7 0
4 9 9 9 3 3 3 3 3 7 7

( scratchpad ) [ { 1 4 3 9 5 3 2 7 0 } reverse [ ] each 4 [ [ dup # " " % ] times drop ] times drop ] " " make print
4 9 9 9 3 3 3 3 3 7 7

I see you like lolis.
who doesn't?

>>8
actually, i can do this instead, just because it looks nicer this way:
{ 1 4 3 9 5 3 2 7 0 } [ reverse [ ] each 4 [ [ dup # " " % ] times drop ] times drop ] " " make print

>>3
that implementation is made of bloated and fail.

'.rleloop'
'lodsw | xchg cx | jcxz .end | lodsw | rep | stosw'
'jmps .rleloop'
'.end'

MOAR!
(wants to see an implementation of this in Lisp or Python).

this 'compression' sucks balls

int compressed_data []={1, 4, 3, 9, 5, 3, 2, 7, 0};
for(int i=0; i<sizeof(compressed_data)/sizeof(int); i+=2)
  while(compressed_data[i]--) std::cout << compressed_data[i+1];

decompress [0]      = []
decompress (x:y:xs) = replicate x y ++ decompress xs

Python, using lists. The final 0 is useless, but I'll respect it. I'll return a list.

Iterative, traditional:

def uncompress(l):
    o = []
    for i in xrange(0, len(l) - 1, 2):
        o += l[i] * [l[i + 1]]
    return o

Iterative, Pythonic:

def uncompress(l):
    o = []
    for x, y in zip(l[:-1:2], l[1:-1:2]):
        o += x * [y]
    return o

Functional, tail recursive:

def uncompress(i, o=[]):
    if i == [0]: return o
    return uncompress2(i[2:], o + i[0] * [i[1]])

(LISP is pretty much the above code, just with a different syntax.)

>>8
who doesn't?
pizza in bizzaro world

>>15 Thank-you Anonymous, I had not seen double :'s in a [ ] or the zip function, but then again I'm not a EXPERT PROGRAMMER.

( scratchpad ) : decompress ( seq -- seq ) [ dup length 2/ [ reverse [ ] each ] swap slip [ [ dup , ] times drop ] times drop ] { } make ;
( scratchpad ) { 1 4 3 9 5 3 2 7 0 } decompress .
{ 4 9 9 9 3 3 3 3 3 7 7 }

>>18
ugly

J ftw.

x =: 1, 4, 3, 9, 5, 3, 2, 7, 0

_2 (< @: ({. # {:) @: >)\ x

BTW, does J have a concatMap function? I couldn't figure out how to join the boxes.

>>18
: decompress ( seq -- seq ) 1 head* 2 group [ reverse >vector dup pop 1- [ dup dup peek swap push ] times >array ] map concat ;


_2 ({. # {:)\ (|: _2 ]\ x)
I actually have no idea what \ does. yay me.

>>22
if you learned J you could understand it

>>23
Well I am learning J. I understand what it does when you give it a vector but in >>22 I gave it a 2xn array and I have no idea how it gave the result.

Superiar.

#/ |: _2 ]\ x