Accounting for no existant characters as inputs C

Question

Sorry if the question title is a little bit off, I had no idea what to call it just because it is such a peculiar question. What I am aiming to do is decode an input string encoded using a method I will explain in a bit, into a plain English text. Encoding is done by choosing an integer nRows between 2 and half the length of the message, e.g. a message of length 11 would allow values of nRows in the range 2 to 5. The message is then written down the columns of a grid, one character in each grid cell, nRows in each column, until all message characters have been used. This may result in the last column being only partially filled. The message is then read out row-wise. For example if the input message was ALL HAIL CAESAR, and the nRows value was 2, encoding would look like this:

A L H I   A S R
L   A L C E A @

Where @ symbolizes a or blank character in the table, that doesn't actually exist - I have simply added it to explain the next part :)

The actual question I have is decoding these phrases. The code I have written thus far works for a few problems, but once the blank characters (@) become many the code begins to break down, as the code obviously does not register them and the algorithm skips past them. My code is:

/*
 * DeConfabulons.c
 * A program to Decode for the Confabulons
 * 
 * August 9th 2015
 */

#include <stdio.h>
#include <string.h>
#include <math.h>

//A simple function confab which given input text encoded using
//the Confabulons encoding scheme, and a number of rows, returns 
//the originally encoded phrase.

void deconfab(const char inText[], int nRows, char outText[])
{
    int count = 0;
    int i = 0;
    int len = strlen(inText);
    float help = ((float)len/(float)nRows);
    int z = 0;
    while (z < round(help))
    {
        while (((int)inText[count] > 0) && (count <= len))
        {
            outText[i] = inText[count]; 
            i ++;
            if (count < (int)help)
            {
                count = count + round((int)help+0.5);
            }
            else
            {
                float helper = count + help;
                count = round(helper);
            }  


        }
        z ++;
        count = z;
    }
    outText[i] = '\0';
}

Which thus far works for the Caesar example I gave earlier. The encoded form of it was ALHI ASRL ALCEA. The main(void) input I have been provided for that problem was:

char buffer[40] = {'\0'};
deconfab("ALHI ASRL ALCEA", 2, buffer);
printf("%s\n", buffer);

Which correctly outputs:

ALL HAIL CAESAR

However when working with cases with extra "blank" characters such as:

char buffer[60] = {0};
char* s = "Two hnvde eo frgqo .uxti hcjeku  mlbparszo y";
deconfab(s, 13, buffer);
printf("%s\n", buffer);

The output should be:

The quick brown fox jumps over the lazy dog.

However my code will return:

Thdefq.the browneorouickmps ov  g x julazy

I have concluded that this caused by the blank characters at the end in the last column by running through multiple tests by hand, however no matter what I try the code will not work for every test case. I am allowed to edit the bulk of the function in nearly any way, however any inputs or anything in int main(void) is not allowed to be edited. I am simply looking for a way to have these blank characters recognized as characters without actually being there (as such) :)

Answer 1

There is some mess with index calculation. At first it is pure discrete transformation. So, it should be implemented using only integer numbers. The function below does what you need.

void deconfab(const char inText[], int nRows, char outText[])
{
    int len = strlen(inText);
    int cols = len / nRows;
    int rows_with_large_cols = len % nRows;
    int count = 0;
    int col = 0;
    int row = 0;

    while (count < len)
    {
        int idx;
        if (row < rows_with_large_cols)
            idx = row * (cols + 1) + col;
        else
            idx = rows_with_large_cols * (cols + 1) +
                (row - rows_with_large_cols) * cols + col;

        if (idx > len - 1) {
            ++col;
            row = 0;
            idx = col;
        }

        outText[count] = inText[idx];
        ++row;
        ++count;
    }
    outText[count] = '\0';
}

It may be rewritten more nicely. Now it is like a pseudocode to explain the algorithm.

Answer 2

First of all, as far as I see, you don't include those "null" characters in your input - if you did that (I guess) by adding any "dummy" characters, the algorithm would work. The reason it does in the first case is that the 'blank' character is missing at the end of the input - the same place as it's missing in the sentence.

You can try to make a workaround by guessing the length of a message with those dummy characters (I'm not sure how to formulate this) like: ALHI ASRL ALCEA has 15 characters (15 mod 2 = 1) but ALHI ASRL ALCEA@ has 16 characters. Similarly, Two hnvde eo frgqo .uxti hcjeku mlbparszo y has 44 characters (44 mod 13 = 5) so you need quite a lot of the dummy chars to make this work (13-5=8).

There are several ways at this point - you can for instance try to insert the missing blank spaces to align the columns, copy everything into a 2-dimensional array char by char, and then read it line by line, or just determine the (len mod rows) characters from the last column, remove them from the input (requires some fiddling with the classic C string functions so I won't give you the full answer here), read the rest and then append the characters from the last column.

I hope this helps.

Answer 3

You cannot use the standard str* functions if you are going to handle nulls. You must, instead, work with the data directly and use the *read family of functions to get your data.