Split string in to array of string in C

Question

I tried to convert a character array to string array. Means single string to multiple sentences.

char str[1000]="This is first. Second sentence is possible. I'm third. The fourth one is here!";

The above string should converted as array of sentences.

char **sentences={
 "This is first.",
 "Second sentence is possible.",
 "I'm third.",
 "The fourth one is here!"
 };

I tried to below code.

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

int
char_comp (char a, char b)
{
  return a == b;
}

int
is_special (char str)
{
  return char_comp (str, '?') || char_comp (str, '.') || char_comp (str, '!');
}

void
str_to_arr (char a[], char **b)
{
  int i = 0, j = 0, length = strlen (a), sentance_length =6, k = 0; //here added sentance_length 8 as static
  for (i = 0; i < sentance_length; i++)
    {
      b[i] = (char *) malloc (length * sizeof (char));
      int n = 0;
      for (j = k; j < length; j++)
    {
      b[i][n] = a[j];

      n++;
      k++;
      if (is_special (a[j]))    //if !,?,. found then move to next sentence.
        {
          b[i][n] = '\0';
          break;
        }

    }
    }
}

int
main ()
{
  char finalstr[1000]; 
    fgets(finalstr, 1000, stdin);
  char **arr = (char **) malloc ((6) * sizeof (char));
  str_to_arr (finalstr, arr);

  for (int i = 0; i < strlen (finalstr); i++)
    {
      printf ("%i--->%s\n",i, arr[i]);
    }
  return 0;
}

When I run the code.

input:

Die schnelle Fourier-Transformation ist ein Algorithmus zur effizienten Berechnung der diskreten Fourier-Transformation.Mit ihr kann ein zeitdiskretes Signal in se ine Frequenzanteile zerlegt und dadurch analysiert werden.  Analog gibt es fuer die diskrete inverse Fourier-Transformation die inverse schnelle Fourier-Transformati on (IFFT). Es kommen bei der IFFT die gleichen Algorithmen, aber mit konjugierten Koeffizienten zur Anwendung. Die FFT gehoert zu den "Divide-and-conquer"-Verfahre n, sodass zuvor berechnete Zwischenergebnisse wiederverwendet werden koennen. Dadu rch koennen arithmetische Rechenoperationen eingespart werden.

output:

0--->��V
1--->Mit ihr kann ein zeitdiskretes Signal in se ine Frequenzanteile zerlegt und dadurch analysiert werden.
2--->  Analog gibt es fuer die diskrete inverse Fourier-Transformation die inverse schnelle Fourier-Transformati on (IFFT).
3---> Es kommen bei der IFFT die gleichen Algorithmen, aber mit konjugierten Koeffizienten zur Anwendung.
4---> Die FFT gehoert zu den "Divide-and-conquer"-Verfahre n, sodass zuvor berechnete Zwischenergebnisse wiederverwendet werden koennen.
5---> Dadu rch koennen arithmetische Rechenoperationen eingespart werden.

input:

This is first. Second sentence is possible. I'm third. The fourth one is here!

output:

0--->This is first.
1---> Second sentence is possible.
2---> I'm third.
3---> The fourth one is here!

When sentences going more than 5. The first sentence becomes ��V like this. What I'm doing wrong here.

Note: I'm not using string.h here. Except strlen.

`int char_comp (char a, char b) { return a == b; }` why do this? — bolov, Dec 15 '21 at 09:39
`char **sentences` Nope. Arrays are not pointers. Pointers are not arrays. — n. m. could be an AI, Dec 15 '21 at 09:43
I am not sure if this is only problem, but it definitely is a problem: Size calculation in `malloc ((6) * sizeof (char))` is using invalid type. — user694733, Dec 15 '21 at 09:43
Look at the accepted answer [here](https://stackoverflow.com/a/2091820/2115408). Just use `"."` as your delimiter. — mmixLinus, Dec 15 '21 at 09:44
@Jabberwocky that's the task. Everything should be done without using string.h — Sarath Kumar, Dec 15 '21 at 09:47
@SarathKumar if you are not allowed to use functions from string.h you need to make that clear in the question. _"Note: I'm not using string.h"_ just means exactly that, no more, no less. — Jabberwocky, Dec 15 '21 at 09:50

score 1 · Accepted Answer · answered Dec 15 '21 at 09:48

char **arr = (char **) malloc ((6) * sizeof (char));

If you are trying to allocate space for 6 sentences then that is wrong. You need to allocate space for the type of elements of arr which are char*, not char:

char **arr = malloc (sizeof *arr * 6);

This allocates only the space for 6 pointers to strings. You now need to allocate the space for each sentence. Something like:

arr[i] = malloc(sizeof *arr * (size_of_sentece + 1))

Also you want to reallocate for each new sentence to not have this upper limit of 6 sentences.

There might be other errors in the code, I haven't checked.

And finally use the standard library and use strtok with ".!?" as delimiters. I understand you can't to this for grades on this task, but nevertheless it's a good thing learn to use the standard library.

Where I can get more info about malloc,realloc for array of string. I tried in the internet couldn't find proper one. — Sarath Kumar, Dec 15 '21 at 10:03
@SarathKumar https://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list — bolov, Dec 15 '21 at 11:16

Split string in to array of string in C

1 Answers1