How to return a matrix of all occurrences of given word/pattern?

Question

I am trying to implement a function that's called Sniffer which gets two inputs and returns the correspond matrix.

First input is an integer binary array values (just zeros or ones)

second input is your searched word ( the word you choose it as argument to your function )

The functionally of the function :

Searching for occurrence of the given word within your given binary array. At each occurrence of your word there's always follows 8 bits following it, assume that always the input is correct (it means that there's no possibility two occurrence occur one after the other without at least there's space 8bits (8 binary values)!).

Then the function must return a matrix(integer matrix) of those followed 8bit for each occurrence of the word sorted corresponded by every row of the matrix. (the functions returns just the first 8bit followed each occurrence of the Searched word)

This means:

First row has first 8 followed bit on first occurrence of the word.
Second row has first 8 followed bit on second occurrence of the word.
Third row has first 8 followed bit on third occurrence of the word.
Fourth row has first 8 followed bit on fourth occurrence of the word.
etc ...

I will elaborate by examples: function structure is Code:

int ** SnifferPattern(int* givenArray , int* SearchedWord);
//it returns the corresponded matrix so used int** ;

example 1:

givenArray = {1,0,1,0,1,1,1,1,1,1,1,1,0,0};
SearchedWord = {1,0,1,0}; 

so the function returns a matrix(size 1x8 - 8 corresponds to 8followed bit) 
the first row is {1,1,1,1,1,1,1,1}, which is the first 8 bit followed 
the word 1010 the matrix here with one row because there's just 
one occurrence of the `SearchedWord` in the given array.

example 2:

givenArray = {1,1,0,0,1,1,1,1,1,1,1,1,1,1,0,0,1,0,1,0,1,0,1,0};
SearchedWord = {1,1,0,0} 

so the function returns a matrix the first row is {1,1,1,1,1,1,1,1} 
which is the first 8 bit followed the word 1010 for the first occurrence. 
for the second occurrence we see the word appear , so the second row of 
the returned matrix(size 2x8) will include the first 8bit followed the 
word 1010 of the second occurrence. so second row of the matrix 
is {1,0,1,0,1,0,1,0} so the returned matrix (size 2x8) has two rows 
(because there's two occurrences of the SearchedWord) which each row 
corresponds to each occurrence of the SearchedWord.

example 3:

givenArray = {1,1,1,0,1,1,1,1,1,1,1,1,0,0}; 
SearchedWord = {0,1,0} 

so the function returns a matrix zero row (like an array with zero values) 
size 1x8 (8 columns is corresponded to 8followed bit). There's no 
occurrence of the SearchedWord within the givenArray so we return a zero 
matrix. There's no overlap between the occurrences of the searchedWords ,
so we assume always correctness of input.

I will explain my algorithm (a pleasure if there's another suggestions more compatible to my case)

My algorithm is searching for the word at every occurrence and then take at every occurrence the first followed 8bit. I take them and store them in a matrix's rows. But it sounds much hard to complete with this.

what I succeeded / tried to implement in C is this:

int ** SnifferPattern(int* s ; int* w)
{
    // s is the given array, w is the searched Word , number occurrences      
    // here 2 , so it should be two prints on the output of the program.
    int n;         
    int a[1000];    
    int i=0;        
    int j;          
    int k = 0;      
    int l;         
    int found = 0;  
    int t = 0;      
 
    a[k++] = i; 
    j = 0;       
    for (i = 0; i < k; i++) 
    {
        n = a[i] - j;   
        if (n == (sizeof(w)/sizeof(w[0]))) 
        {
            t = 0;
            for (l = 0; w[l]; l++)  
            {
                if (s[l + j] == w[l])
                {
                    t++;  // Matched a character.
                }
            }
            
            if (t == (sizeof(w)/sizeof(w[0]))) 
            {
                found++;  // We found a match!
                printf("word occurred at location=%d \n", j);  // Pint location
            }
        }
        j = a[i] + 1; 
    }
}

int main() {

    int s[1000] = {1,0,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1};
    int w[1000] = {1,0,1}; 
    
    // s is the given array, w is the searched Word , number occurrences      
    // here 2 , so it should be two prints on the output of the program.
    
    SnifferPattern(s , w)
    
   //I should print all the matrix's row in the main function .
    return 0;
}

Answer 1

I think I have figured out what you need. And, as stated in your question ("at each occurrence of your word there's always follows 8 bits"), the following requires that a least 8-integers follow any match of w in s. Since you will include 8-integers in each row of the matrix you return, then using a pointer-to-array-of int[8] allows a single free() of the result in the caller.

Your sniffer function will loop over each integer in s keeping a counter index (ndx) of each time an integer in w matches the integer in s. When ndx equals the number of elements in w a match has been found and the next 8 integers are are collected as the columns in that row of your matrix using next8 as the index. You could write your sniffer function as:

#define AFTER    8

/* function returning allocated pointer to array of 8 int, *n of them */
int (*sniffer (int *s, int selem, int *w, int welem, int *n))[AFTER]
{
    int ndx = 0,                    /* match index */
        next8 = AFTER,              /* counter for next 8 after match found */
        found = 0,                  /* number of tiems w found in s */
        (*matrix)[AFTER] = NULL;
    
    for (int i = 0; i < selem; i++) {           /* loop over each int in s */
        if (next8 < AFTER) {                    /* count if next8 less than AFTER */
            matrix[found-1][next8++] = s[i];    /* set next8 index in matrix row */
        }
        else if (s[i] == w[ndx]) {              /* if s[i] matches w[ndx] */
            if (++ndx == welem) {               /* increment ndx, compare w/welem */
                /* allocate storage for next row in matrix */
                if (!(matrix = realloc (matrix, (found + 1) * sizeof *matrix))) {
                    perror ("realloc-matrix");  /* on failure, handle error */
                    *n = 0;
                    return NULL;
                }
                /* (optional) set all elements in row 0 */
                memset (matrix[found], 0, sizeof *matrix);
                found += 1;                     /* increment found count */
                ndx = next8 = 0;                /* reset ndx, set next8 to count */
            }
        }
        else            /* otherwise, not a match and not counting */
            ndx = 0;    /* set match index to 0 */
    }
    
    *n = found;         /* update value at n to found, so n available in main */
    
    return matrix;      /* return allocated matrix */
}

(note: the number of elements in s is provided in selem and the number of elements in w is provided by welem)

Changing your s[] in main to int s[] = {1,0,1,0,0,0,0,1,1,1,1,1,0,1,1,1,1,1,0,0,0,0} so it is easy to verify the results, you could write you program as:

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

#define ARSZ  1000      /* if you need a constant, #define one (or more) */
#define AFTER    8

/* function returning allocated pointer to array of 8 int, *n of them */
int (*sniffer (int *s, int selem, int *w, int welem, int *n))[AFTER]
{
    int ndx = 0,                    /* match index */
        next8 = AFTER,              /* counter for next 8 after match found */
        found = 0,                  /* number of tiems w found in s */
        (*matrix)[AFTER] = NULL;
    
    for (int i = 0; i < selem; i++) {           /* loop over each int in s */
        if (next8 < AFTER) {                    /* count if next8 less than AFTER */
            matrix[found-1][next8++] = s[i];    /* set next8 index in matrix row */
        }
        else if (s[i] == w[ndx]) {              /* if s[i] matches w[ndx] */
            if (++ndx == welem) {               /* increment ndx, compare w/welem */
                /* allocate storage for next row in matrix */
                if (!(matrix = realloc (matrix, (found + 1) * sizeof *matrix))) {
                    perror ("realloc-matrix");  /* on failure, handle error */
                    *n = 0;
                    return NULL;
                }
                /* (optional) set all elements in row 0 */
                memset (matrix[found], 0, sizeof *matrix);
                found += 1;                     /* increment found count */
                ndx = next8 = 0;                /* reset ndx, set next8 to count */
            }
        }
        else            /* otherwise, not a match and not counting */
            ndx = 0;    /* set match index to 0 */
    }
    
    *n = found;         /* update value at n to found, so n available in main */
    
    return matrix;      /* return allocated matrix */
}

int main (void) {
    
    int s[] = {1,0,1,0,0,0,0,1,1,1,1,1,0,1,1,1,1,1,0,0,0,0},
        w[] = {1,0,1},
        n = 0,
        (*result)[AFTER] = NULL;
    
    result = sniffer (s, sizeof s/sizeof *s, w, sizeof w/sizeof *w, &n);
    
    for (int i = 0; i < n; i++) {                       /* loop over result matrix */
        printf ("matrix[%d] = {", i);                   /* output prefix */
        for (int j = 0; j < AFTER; j++)                 /* loop over column values */
            printf (j ? ",%d" : "%d", result[i][j]);    /* output column value  */
        puts ("}");                                     /* output suffix and \n */
    }
    
    free (result);      /* free allocated memory */
}

Example Use/Output

$ ./bin/pattern_in_s
matrix[0] = {0,0,0,0,1,1,1,1}
matrix[1] = {1,1,1,1,0,0,0,0}

If I have misunderstood your question, please let me know in a comment below and I'm happy to help further. Let me know if you have any questions.

Answer 2

There are several issues you must solve.

How are arrays represented?

You have arrays of integers, whose valid values can be 0 or 1. How do you datermine the length of sich an array. There are basically two possibilities:

• Use a sentinel value. That's how C strings are stored: The actual string is terminated by the special value '\0', the null terminator. The memory used to store the string may be larger. In your case, you could use a value that isn't used:

  enum {
      END = -1
  };
  
  int a[] = {0, 1, 0, 1, END};
  

  The disadvantage here is that you must be careful not to forget the explicit terminator. Also, if you want to find out the length of an array, you must walk it to the end. (But that's not an issue with small arrays.)

• Use an explicit length that goes along with the array.

  int a[] = {1, 0, 1, 0};
  int alen = sizeof(a) / sizeof(*a);
  

  The disadvantage here is that you must pass the length to any function that operates on the array. The function f(a) does not know kow long a is; the function should be something like f(a, alen). (The sizeof(a) / sizeof(*a) mechanism works only when the array a is in scope. See here for a discussion on how to find the length of an array.)

  You could, of course, define a struct that combines data and length.

How do you return an array from a function?

That's your actual question. Again there are several possibilities:

• Return the array. That usually means to allocate the array you want to return on the heap with malloc, which means that the caller must call free on the result at some time. The caller must know how big the returned array is. You can use sentinels as described above or you could pass in a pointer to a sive variable, which the function fills in:

  int **f(..., int *length) { ... }
  

  Call this function like this:

  int length;
  int **p = f(..., &length);
  
  for (int i = 0; i < length; i++) ...
  

• Pass in an array and hve the function fill it. That means that the function must know about the size of the array. The return value can be used to return the actual size of the array:

  int f(..., int **res, int max) { ... }
  

  Call this function like this:

  int *p[20];
  int length = f(..., p, 20);
  
  for (int i = 0; i < length; i++) ...
  

Let's apply this to your problem.

You want to match a pattern in a string and then return a list of all 8-bit sequences after the matches. Let's represent an array as array + length. Your function might then look like this:

int sniffer(const int *s, int slen,         // haystack array + length
            const int *w, int wlen,         // needle array + length
            const int **res, int reslen)    // results array
{ ... }

It passes in the arrays s and w plus their lengths. It also passes in a third array plus its length. That array hold the results. The number of valid results – the number of rows in your matrix – is the returned value.

Call this function:

int s[] = {...};
int w[] = {...};

const int *res[8];
int n = sniffer(s, sizeof(s) / sizeof(*s),
                w, sizeof(w) / sizeof(*w),
                res, sizeof(res) / sizeof(*res));                    

What happens if there are more than reslen matches? The excess matches cannot be written, of course, but do they contribute to the return value? If they do, you could pass in an array length of 0 just to see how many matches there are. (That' s what the string function snprintf does, in a way.) If they don't you get the exact length of the result array. Both strategies are valid.

Here's an example implementation. It uses your test case #2:

#include <stdlib.h>
#include <stdio.h>

/* 
 *  test whether the next len elements of s and w match
 */
int match(const int *s, const int *w, int len)
{
    while (len-- > 0) {
        if (*s++ != *w++) return 0;
    }
    
    return 1;
}

int sniffer(const int *s, int slen,         // haystack array + length
            const int *w, int wlen,         // needle array + length
            const int **res, int reslen)    // results array
{
    int n = 0;
    
    for (int i = 0; i <= slen - wlen - 8; i++) {
        const int *p = s + i;
    
        if (match(p, w, wlen)) {
            if (n < reslen) res[n] = p + wlen;
            n++;
        }
    }  
  
    return n;
}



int main(void)
{
    int s[] = {1, 1, 0, 0, 1, 1, 1, 1,
               1, 1, 1, 1, 1, 1, 0, 0,
               1, 0, 1, 0, 1, 0, 1, 0};
    int w[] = {1, 1, 0, 0};

    const int *res[8];
    int reslen = sizeof(res) / sizeof(*res);
    
    int n = sniffer(s, sizeof(s) / sizeof(*s),
                    w, sizeof(w) / sizeof(*w),
                    res, reslen);                    

    printf("%d patterns:\n", n);
    for (int i = 0; i < n && i < reslen; i++) {
        printf("[%d] {", i);
        
        for (int j = 0; j < 8; j++) {
            if (j) printf(", ");
            printf("%d", res[i][j]);
        }
        
        printf("}\n");
    }

    return 0;
}