Thread: RLE Encoding/Decoding using simple C concept..plz help

> Output for above input: 1\12\13\34\25\5A\3
How do you propose to tell the difference at the start between
1 repeated once, followed by 2, as in 12
1 repeated 12 times followed by a \, as in 111111111111\

The problem is, your encoding is wrong.
You should be writing out just TWO bytes per character, one byte being the character, and the other byte representing the repeat count.

Decoding is them much easier as you always read a pair, then expand a sequence.

and you should work on your indentation - it is not consistent

Agreed, the encoder is completely wrong, in so many ways!
But since it's urgent you wont have time to understand the details so I wont bother to explain them.

His encoding, if really done in text, could work. (It could be decoded.) Example,

Code:

1/12/1

Read the 1. Read the /. Now read in numbers until you get to the next /. That last number/character was just read, put it back. You now have: character / number of times to repeat.

Granted, this encoding, if done in plain text, is horrid, and I wonder just how much benefit you'd get out of it. It should really be a binary file of two bytes: character, length. However, whatever requirements are being put on the OP may restrict that.

EDIT: Rereading the OP's first post, I think perhaps whatever he or she has is trying to say, put it in a binary file, with byte pairs of character,length... albeit the whole thing about escape sequences is ... off.

> but the program is running perfectly..
Just because it runs without crashing and produces an output file is not a sign of perfection.

You're not addressing the issues raised in my previous post.

Given 1/122/3
You need to find a way of making sure it decodes as
1/122/3
and not as
1/122/3
This is what Cactus_Hugger was talking about.

It might be easier if you always wrote the length out as 3 digits, say
1/0122/003
If you know the length is always 3 digits, it makes it easier to deal with.

Your description clearly indicates that you are not supposed to store character followed by backslash and integer, but a character followed by another character whose integral value indicates the number of repeats.

So your output should look something like that:

A0B"C^
for 48 A-s, 34 B-s, 94 C-s

The description uses \1 and explains how to interpret it, because characters with small ASCII values are non-printing (so you'd just see rectangles instead, for example).

A character is an integer, just with a small range. You don't want to print the integral value of the character at all. You want to use a char for counting, not an int.

Code:

char a = 's';
char b;

for (b = 0; b < a; ++b) {
}

Try something like that putting something inside the loop, and you might get the idea.

You might also Google for ASCII tables.

Originally Posted by garv84

plz tell me something..

say for ex : 1\125

all thses are characters.we are to print 1 125 tyms.how can we conver characters 125 into integar..??

If you write the compressor properly then you don't have to!

Everyone always teaches RLE compression by giving examples with text, and shows the compressed version as if it were text. Such examples always contain lots of repeated bytes too.
However that's not the reality of how anybody who knows what they're doing implements it.

You are supposed to accept binary data, and write out binary data. Every byte value between 0 and 255 should be able to be compressed from the input file, and later restored after decompression of hte compressed file. This in turn basically means that your compressed file becomes a binary format as well.

... and shows the compressed version as if it were text. Such examples always contain lots of repeated bytes too.

You could of course use RLE several times over to compress you data virtually to nothing.

Just kidding.