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) :)
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.
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.
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.
Related
I wrote a program that counts and prints the number of occurrences of elements in a string but it throws a garbage value when i use fgets() but for gets() it's not so.
Here is my code:
#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<stdlib.h>
int main() {
char c[1005];
fgets(c, 1005, stdin);
int cnt[26] = {0};
for (int i = 0; i < strlen(c); i++) {
cnt[c[i] - 'a']++;
}
for (int i = 0; i < strlen(c); i++) {
if(cnt[c[i]-'a'] != 0) {
printf("%c %d\n", c[i], cnt[c[i] - 'a']);
cnt[c[i] - 'a'] = 0;
}
}
return 0;
}
This is what I get when I use fgets():
baaaabca
b 2
a 5
c 1
32767
--------------------------------
Process exited after 8.61 seconds with return value 0
Press any key to continue . . . _
I fixed it by using gets and got the correct result but i still don't understand why fgets() gives wrong result
Hurray! So, the most important reason your code is failing is that your code does not observe the following inviolable advice:
Always sanitize your inputs
What this means is that if you let the user input anything then he/she/it can break your code. This is a major, common source of problems in all areas of computer science. It is so well known that a NASA engineer has given us the tale of Little Bobby Tables:
Exploits of a Mom #xkcd.com
It is always worth reading the explanation even if you get it already #explainxkcd.com
medium.com wrote an article about “How Little Bobby Tables Ruined the Internet”
Heck, Bobby’s even got his own website — bobby-tables.com
Okay, so, all that stuff is about SQL injection, but the point is, validate your input before blithely using it. There are many, many examples of C programs that fail because they do not carefully manage input. One of the most recent and widely known is the Heartbleed Bug.
For more fun side reading, here is a superlatively-titled list of “The 10 Worst Programming Mistakes In History” #makeuseof.com — a good number of which were caused by failure to process bad input!
Academia, methinks, often fails students by not having an entire course on just input processing. Instead we tend to pretend that the issue will be later understood and handled — code in academia, science, online competition forums, etc, often assumes valid input!
Where your code went wrong
Using gets() is dangerous because it does not stop reading and storing input as long as the user is supplying it. It has created so many software vulnerabilities that the C Standard has (at long last) officially removed it from C. SO actually has an excellent post on it: Why is the gets function so dangerous that it should not be used?
But it does remove the Enter key from the end of the user’s input!
fgets(), in contrast, stops reading input at some point! However, it also lets you know whether you actually got an entire line of of text by not removing that Enter key.
Hence, assuming the user types: b a n a n a Enter
gets() returns the string "banana"
fgets() returns the string "banana\n"
That newline character '\n' (what you get when the user presses the Enter key) messes up your code because your code only accepts (or works correctly given) minuscule alphabet letters!
The Fix
The fix is to reject anything that your algorithm does not like. The easiest way to recognize “good” input is to have a list of it:
// Here is a complete list of VALID INPUTS that we can histogram
//
const char letters[] = "abcdefghijklmnopqrstuvwxyz";
Now we want to create a mapping from each letter in letters[] to an array of integers (its name doesn’t matter, but we’re calling it count[]). Let’s wrap that up in a little function:
// Here is our mapping of letters[] ←→ integers[]
// • supply a valid input → get an integer unique to that specific input
// • supply an invalid input → get an integer shared with ALL invalid input
//
int * histogram(char c) {
static int fooey; // number of invalid inputs
static int count[sizeof(letters)] = {0}; // numbers of each valid input 'a'..'z'
const char * p = strchr(letters, c); // find the valid input, else NULL
if (p) {
int index = p - letters; // 'a'=0, 'b'=1, ... (same order as in letters[])
return &count[index]; // VALID INPUT → the corresponding integer in count[]
}
else return &fooey; // INVALID INPUT → returns a dummy integer
}
For the more astute among you, this is rather verbose: we can totally get rid of those fooey and index variables.
“Okay, okay, that’s some pretty fancy stuff there, mister. I’m a bloomin’ beginner. What about me, huh?”
Easy. Just check that your character is in range:
int * histogram(char c) {
static int fooey = 0;
static int count[26] = {0};
if (('a' <= c) && (c <= 'z')) return &count[c - 'a'];
return &fooey;
}
“But EBCDIC...!”
Fine. The following will work with both EBCDIC and ASCII:
int * histogram(char c) {
static int fooey = 0;
static int count[26] = {0};
if (('a' <= c) && (c <= 'i')) return &count[ 0 + c - 'a'];
if (('j' <= c) && (c <= 'r')) return &count[ 9 + c - 'j'];
if (('s' <= c) && (c <= 'z')) return &count[18 + c - 's'];
return &fooey;
}
You will honestly never have to worry about any other character encoding for the Latin minuscules 'a'..'z'.Prove me wrong.
Back to main()
Before we forget, stick the required magic at the top of your program:
#include <stdio.h>
#include <string.h>
Now we can put our fancy-pants histogram mapping to use, without the possibility of undefined behavior due to bad input.
int main() {
// Ask for and get user input
char s[1005];
printf("s? ");
fgets(s, 1005, stdin);
// Histogram the input
for (int i = 0; i < strlen(s); i++) {
*histogram(s[i]) += 1;
}
// Print out the histogram, not printing zeros
for (int i = 0; i < strlen(letters); i++) {
if (*histogram(letters[i])) {
printf("%c %d\n", letters[i], *histogram(letters[i]));
}
}
return 0;
}
We make sure to read and store no more than 1004 characters (plus the terminating nul), and we prevent unwanted input from indexing outside of our histogram’s count[] array! Win-win!
s? a - ba na na !
a 4
b 1
n 2
But wait, there’s more!
We can totally reuse our histogram. Check out this little function:
// Reset the histogram to all zeros
//
void clear_histogram(void) {
for (const char * p = letters; *p; p++)
*histogram(*p) = 0;
}
All this stuff is not obvious. User input is hard. But you will find that it doesn’t have to be impossibly difficult genius-level stuff. It should be entertaining!
Other ways you could handle input is to transform things into acceptable values. For example you can use tolower() to convert any majuscule letters to your histogram’s input set.
s? ba na NA!
a 3
b 1
n 2
But I digress again...
Hang in there!
I have an assignment that basically is asking to justify a paragraph given line length. So for instance the paragraph
"I am a student of C, this is my first assignment. I hope I finish on time." given line length of 17 should be as follows:
output
I am a student of
C, this is my
first assignment.
I hope I finish
on time.
I am having trouble with dynamically placing spacing in between the words. I currently have a function that counts the words in a paragraph and stores them into a 2d array but I have no idea how to a) calculate the amount of spacing in between words and b) how to dynamically print that justified paragraph.
Here is the code I have so far:
int getAllWordsFrom2DArray(char *paragraph, char words[MAX_NUMBER_OF_WORDS][MAX_WORD_LENGTH]) {
int i,j,totalWords = 0;
for(i=0; i < strlen(paragraph); i++) {
int wordLength;
if (paragraph[i] == ' ' || paragraph[i+1] == '\0') {
totalWords++;
wordLength = i;
for(j=0; j < wordLength; j++) {
words[i][j] = paragraph[j];
}
}
}
printf("%s", words);
return totalWords;
}
//Code in progress
int getNumberOfWordsForNextLine(int totalWords, int lineLength, char words[MAX_NUMBER_OF_WORDS][MAX_WORD_LENGTH]) {
int wordsForNextLine = 0;
for(int i=0; i < totalWords; i++) {
wordsForNextLine = 0 ;
}
}
//code in progress
void printNextLine(int wordsForNextLine) {
}
//skeleton code provided by instructor
void justifyAndPrintParagraph(char* paragraph, int lineLength) {
char words[MAX_NUMBER_OF_WORDS][MAX_WORD_LENGTH];
int totalWords = getAllWordsFrom2DArray(paragraph, words);
int processedWords = 0;
while (processedWords < totalWords) {
int wordsForNextLine = getNumberOfWordsForNextLine(totalWords, lineLength, words);
printNextLine(wordsForNextLine);
processedWords += wordsForNextLine;
}
}
To clarify, we are not allowed to use strlok. Essentially we are expected to just use the basics in doing this. I need to use the void justifyAndPrintParagraph function and signature but other than that I'm free to do whatever.
Edit: I forgot to add that if spaces cannot be evenly divided then the extra spaces are to be allocated left to right.
Any help is greatly appreciated.
Consider how many spaces you have to distribute. For example, given the input:
18
I am the very model of a modern Major-General.
Computing the number of words that fit on the line goes:
"I" + "am" + "the" + "very" + (4-1 words) --> 13
"I" + "am" + "the" + "very" + "model" + (5-1 words) --> 19
So only the first 4 words fit on an 18-character line. The number of space characters to distribute are then easily calculated:
N = max_line_width - sum_of_word_lengths
Now for the hard part: how many spaces between each word? Your homework expects you to divvy extra unbalanced spaces left-to-right, meaning that each pair of words may have a different number of space characters.
However, the difference will always be a single space character. Take a moment to convince yourself this is true:
I···am···the··very
-2-4-6-8-0-2-4-6-8
In our little example, we find that there are three space characters in the first two inter-word spacings, and two space characters in the last.
The minimum number of space characters per inter-word spacing is easy enough to caluclate:
nsp = N / (number_of_words_in_line - 1)
Beware! What happens if you have only one word on the line? (Do you really need to distribute spaces for such a line?)
And now, for the cool tricky math part, you can calculate the number of times you need to add a space to the inter-word spacing as:
nplus1 = N - nsp * (number_of_words_in_line - 1)
or just:
nplus1 = N % (number_of_words_in_line - 1)
Keep in mind that it is possible that all inter-word spacings are the same number of space characters, and may be exactly one space character even. Notice how our calculations work just as well in those cases.
Now you can print the words for the line in a loop, adding nsp space characters after every word, plus an extra space after the first nplus1 words.
Remember, the last word of the line doesn’t get any spaces. It is followed by a newline!
Hopefully this should help you work your way through this assignment.
(I personally think it is a bit of a careless assignment as your first ever, introduction to C class.)
And now, if I have made errors, it is because I am very, very sleepy. Someone will surely point it out if I have.
So using Dúthomhas' suggestion I was able to create the function below:
void justifyAndPrintLine(char words[MAX_NUMBER_OF_WORDS][MAX_WORD_LENGTH], int processedWords, int amountOfWordsForNextLine, int lineLength) {
int total = 0;
for (int i = processedWords; i < processedWords + amountOfWordsForNextLine; i++) {
total += (int) strlen(words[i]);
}
int spaces = lineLength - total;
int spacesBetweenWords = spaces / (amountOfWordsForNextLine - 1);
int spacesRemaining = spaces % (amountOfWordsForNextLine - 1);
int spaceForThisWord;
int leftWords = processedWords + amountOfWordsForNextLine;
while (processedWords != leftWords) {
spaceForThisWord = spacesBetweenWords;
if (spacesRemaining > 0) {
spaceForThisWord++;
spacesRemaining--;
}
printLine(words[processedWords], spaceForThisWord);
processedWords++;
}
}
A key part of my understanding of the math was that the difference in spacing was always going to a single space character. Borrowing his math I was able to properly justify the paragraph. Thanks again Dúthomhas!
I am trying to print a 2D array that should look like this
*************
*
*
*************
like a frame, but I have hard time because it is not printing the stars instead it prints numbers and even those columns I specified with 42 ('*') are not showing. I don't know if the problem is the way I am printing the array. I hope someone can help me. Thank you!
here's my code: int i,j;
int star=42;
char starchar=(char)star;
//colMax=10;
int frame[rowMax][colMax];
for(row=0;row<rowMax;row++)
{
for(column=0;column<colMax;column++)
{
frame[0][column]=starchar;
frame[9][column]=starchar;
frame[row][9]=starchar;
printf("%d%d",row,column);
}
It is printing numbers because you are passing %d%d to the printf function, which stands for decimal number. Replace %d with %c to get the character representation. See the manual for printf, e.g. at http://www.cplusplus.com/reference/cstdio/printf/
Also, you have another problem which is you are passing values other than the ones you intend into printf. You are passing row and column, which are not the values at that location, but the numbers of the index. You want something like:
printf("%c",frame[row][column]);
As Paul Roub points out in comments, you are also doing some odd things with the way you are putting values into the array during your display loop, which is probably also going to confuse you if you develop this code further. First, you don't ever specify what is in the non-star parts of the array. Those should be set to something like the space character before, or at the same time, as adding stars. As written, the values will default to zero, which is unprintable null in ASCII, not space (which is 32).
Also, I assume you mean to be setting up the values first, and then displaying them. The way your loop works, it is setting the values as it draws them, in an interesting way that I would assume is not what you intend. Your example will currently end up showing what you expect, but usually I would expect you to fill in the values first and then display them later.
printf("%d%d",row,column); prints the row and column, not the value in frame[row][column].
#include <string.h>
#include <stdio.h>
int main(void) {
// Not clear if rowMax is the maximum index or the width. Assume maximum index.
size_t rowMax = 10;
size_t colMax = 5;
int star = '*';
// Much easier to use char array
char frame[rowMax + 1][colMax + 1]; // +1 the dimensions
// fill
memset(&frame[0], star, colMax + 1); // fill first row
for (size_t row = 1; row < rowMax; row++) {
memset(&frame[row], ' ', colMax); // fill with spaces
frame[row][colMax] = star;
}
memset(&frame[rowMax], star, colMax + 1); // fill last row
// print
for (size_t row = 0; row <= rowMax; row++) {
for (size_t col = 0; col <= colMax; col++) {
fputc(frame[row][col], stdout);
}
fputc('\n', stdout);
}
return 0;
}
I'm trying to implement Boyer-Moore Algorithm in C for searching a particular word in .pcap file. I have referenced code from http://ideone.com/FhJok5. I'm using this code as it is.
Just I'm passing packet as string and the keyword I'm searching for to the function search() in it. When I'm running my code it is giving different values every time. Some times its giving correct value too. But most of times its not identifying some values.
I have obtained results from Naive Algo Implementation. Results are always perfect.
I am using Ubuntu 12.0.4 over VMware 10.0.1. lang: C
My question is It has to give the same result every time right? whether right or wrong. This output keeps on changing every time i run the file on same inputs; and during several runs, it gives correct answer too. Mostly the value is varying between 3 or 4 values.
For Debugging I did so far:
passed strings in stead of packet every time, Its working perfect and same and correct value every time.
checking pcap part, I can see all packets are being passed to the function (I checked by printing packet frame no).
same packets I am sending to Naive Algo code, its giving perfect code.
Please give me some idea, what can be the issue. I suspect some thing wrong with memory management. but how to find which one?
Thanks in advance.
# include <limits.h>
# include <string.h>
# include <stdio.h>
# define NO_OF_CHARS 256
// A utility function to get maximum of two integers
int max (int a, int b) { return (a > b)? a: b; }
// The preprocessing function for Boyer Moore's bad character heuristic
void badCharHeuristic( char *str, int size, int badchar[NO_OF_CHARS])
{
int i;
// Initialize all occurrences as -1
for (i = 0; i < NO_OF_CHARS; i++)
badchar[i] = -1;
// Fill the actual value of last occurrence of a character
for (i = 0; i < size; i++)
badchar[(int) str[i]] = i;
}
/* A pattern searching function that uses Bad Character Heuristic of
Boyer Moore Algorithm */
void search( char *txt, char *pat)
{
int m = strlen(pat);
int n = strlen(txt);
int badchar[NO_OF_CHARS];
/* Fill the bad character array by calling the preprocessing
function badCharHeuristic() for given pattern */
badCharHeuristic(pat, m, badchar);
int s = 0; // s is shift of the pattern with respect to text
while(s <= (n - m))
{
int j = m-1;
/* Keep reducing index j of pattern while characters of
pattern and text are matching at this shift s */
while(j >= 0 && pat[j] == txt[s+j])
j--;
/* If the pattern is present at current shift, then index j
will become -1 after the above loop */
if (j < 0)
{
printf("\n pattern occurs at shift = %d", s);
/* Shift the pattern so that the next character in text
aligns with the last occurrence of it in pattern.
The condition s+m < n is necessary for the case when
pattern occurs at the end of text */
s += (s+m < n)? m-badchar[txt[s+m]] : 1;
}
else
/* Shift the pattern so that the bad character in text
aligns with the last occurrence of it in pattern. The
max function is used to make sure that we get a positive
shift. We may get a negative shift if the last occurrence
of bad character in pattern is on the right side of the
current character. */
s += max(1, j - badchar[txt[s+j]]);
}
}
/* Driver program to test above function */
int main()
{
char txt[] = "ABAAAABAACD";
char pat[] = "AA";
search(txt, pat);
return 0;
This problem has been irritating me for too long. I need a non-recursive algorithm in C to generate non-distinct character strings. For instance, if a given character string is 26 characters long, and the string is of length 2, then there are 26^2 non-distinct characters.
Please note that these are distinct combinations, aab is not the same as baa or aba. I've searched S.O., and most solutions produce non-distinct combinations. Also, I do not need permutations.
The algorithm can't rely on a libraries. I'm going to translate this C code into cuda where standard C libraries don't work (at least not efficiently).
Before I show you what I started, let me explain an aspect of the program. It is multithreaded on a GPU, so I initialize the beginning string with a few characters, aa in this case. To create a combination, I add one or more characters depending on the desired length.
Here's one method that I have attempted:
int main(void){
//Declarations
char final[12] = {0};
char b[3] = "aa";
char charSet[27] = "abcdefghijklmnopqrstuvwxyz";
int max = 4; //Set for demonstration purposes
int ul = 1;
int k,i;
//This program is multithreaded on a GPU. Each thread is initialized
//to a starting value for the string. In this case, it is aa
//Set final with a starting prefix
int pref = strlen(b);
memcpy(final, b, pref+1);
//Determine the number of non-distinct combinations
for(int j = 0; j < length; j++) ul *= strlen(charSet);
//Start concatenating characters to the current character string
for(k = 0; k < ul; k++)
{
final[pref+1] = charSet[k];
//Do some work with the string
}
...
It should be obvious that this program does nothing useful, accept if I'm only appending one character from charSet.
My professor suggested that I try using a mapping (this isn't homework; I asked him about possible ways to generate distinct combinations without recursion).
His suggestion is similar to what I started above. Using the number of combinations calculated, he suggested to decompose it according to mod 10. However, I realized it wouldn't work.
For example, say I need to append two characters. This gives me 676 combinations using the character set above. If I am on the 523rd combination, the decomposition he demonstrated would yield
523 % 10 = 3
52 % 10 = 2
5 % 10 = 5
It should be obvious that this doesn't work. For one, it yields three characters, and two, if my character set is larger than 10 characters, the mapping ignores those above index 9.
Still, I believe a mapping is key to the solution.
The other method I explored utilized for loops:
//Psuedocode
c = charset;
for(i = 0; i <length(charset); i++){
concat string
for(j = 0; i <length(charset); i++){
concat string
for...
However, this hardcodes the length of the string I want to compute. I could use an if statement with a goto to break it, but I would like to avoid this method.
Any constructive input is appreciated.
Given a string, to find the next possible string in the sequence:
Find the last character in the string which is not the last character in the alphabet.
Replace it with the next character in the alphabet.
Change every character to the right of that character with the first character in the alphabet.
Start with a string which is a repetition of the first character of the alphabet. When step 1 fails (because the string is all the last character of the alphabet) then you're done.
Example: the alphabet is "ajxz".
Start with aaaa.
First iteration: the rightmost character which is not z is the last one. Change it to the next character: aaaj
Second iteration. Ditto. aaax
Third iteration: Again. aaaz
Four iteration: Now the rightmost non-z character is the second last one. Advance it and change all characters to the right to a: aaja
Etc.
First, thanks for everyone's input; it was helpful. Being that I am translating this algorithm into cuda, I need it to be as efficient as possible on a GPU. The methods proposed certainly work, but not necessarily optimal for GPU architecture. I came up with a different solution using modular arithmetic that takes advantage of the base of my character set. Here's an example program, primarily in C with a mix of C++ for output, and it's fairly fast.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
typedef unsigned long long ull;
int main(void){
//Declarations
int init = 2;
char final[12] = {'a', 'a'};
char charSet[27] = "abcdefghijklmnopqrstuvwxyz";
ull max = 2; //Modify as need be
int base = strlen(charSet);
int placeHolder; //Maps to character in charset (result of %)
ull quotient; //Quotient after division by base
ull nComb = 1;
char comb[max+1]; //Array to hold combinations
int c = 0;
ull i,j;
//Compute the number of distinct combinations ((size of charset)^length)
for(j = 0; j < max; j++) nComb *= strlen(charSet);
//Begin computing combinations
for(i = 0; i < nComb; i++){
quotient = i;
for(j = 0; j < max; j++){ //No need to check whether the quotient is zero
placeHolder = quotient % base;
final[init+j] = charSet[placeHolder]; //Copy the indicated character
quotient /= base; //Divide the number by its base to calculate the next character
}
string str(final);
c++;
//Print combinations
cout << final << "\n";
}
cout << "\n\n" << c << " combinations calculated";
getchar();
}