Split a string into an array of strings

Question

It's been a while that I haven't written programs in C. I am used to write code in C#.

So, I want to split the user string input into an array of strings using a delimiter. I did this, but when I want to fetch the array I have a segmentation fault. As an example, I want just to print one element of the array.

I have checked on the net, but nothing worked.

Any hints ?

Thanks

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

int main ()
{

  char function[] = {};
  char * pch;
  int cpt = 0;
  int nb_terms = 0;

  printf("Entrez le nombre de termes :\n");
  scanf("%d", &nb_terms);

  char word[nb_terms];

  printf("Entrez votre fonction en n'utilisant que les 5 caractères suivants (a,b,c,d et +) :\n");
  scanf("%s", &function);

  pch = strtok (function,"+");
  while (pch != NULL)
  {
    word[cpt++] = pch;
    printf ("%s\n",pch);
    pch = strtok (NULL, "+");
  }

  printf ("%s\n",&word[1]);

  return 0;

}

Could you give an example of your input and expected output? — Schwern
– Schwern, Commented May 30, 2018 at 23:23
@wildplasser It's a GNU extension for a zero size array. gcc.gnu.org/onlinedocs/gcc/Zero-Length.html — Schwern
– Schwern, Commented May 30, 2018 at 23:24
You're getting the Segmentation fault because the array function is of a fixed size of none. Use a pointer to a char instead and allocate with malloc or establish a big enough size for the array. — Kolt Penny
– Kolt Penny, Commented May 30, 2018 at 23:26
warning are there not because the compiler writers hate you , but because they are trying to help — pm100
– pm100, Commented May 30, 2018 at 23:44

Schwern · Accepted Answer · 2018-05-30 23:42:05Z

Compiler warnings reveal your problems.

cc -Wall -Wshadow -Wwrite-strings -Wextra -Wconversion -std=c99 -pedantic -g   -c -o test.o test.c
test.c:7:21: warning: use of GNU empty initializer extension [-Wgnu-empty-initializer]
  char function[] = {};
                    ^
test.c:7:21: warning: zero size arrays are an extension [-Wzero-length-array]
test.c:18:15: warning: format specifies type 'char *' but the argument has type 'char (*)[0]'
      [-Wformat]
  scanf("%s", &function);
         ~~   ^~~~~~~~~

These are all related. char function[] = {} is a GNU extension to declare a 0 size array. Then you try to put stuff into it, but it's size 0. So there's going to be an overflow.

Instead, you need to allocate some space to function and be sure to limit scanf to only that size, no larger.

// {0} initializes all characters to 0.
// 1024 is a good size for a static input buffer.
char function[1024] = {0};

// one less because of the null byte
scanf("%1023s", &function);

The next warning...

test.c:23:17: warning: incompatible pointer to integer conversion assigning to 'char' from 'char *';
      dereference with * [-Wint-conversion]
    word[cpt++] = pch;
                ^ ~~~
                  *

is because you're trying to put the string (char *) pch where a character (char) goes. Even if you're only reading a single character from strtok (which you cannot guarantee) it always returns a string. You want an array of strings (char **). It also helps to have descriptive variable names.

char *word;                 // this was pch
char *words[nb_terms];      // this was word

After changing pch to word, and word to words in the rest of the code it all works.

  size_t word_idx = 0;
  for(
      char *word = strtok(function,"+");
      word != NULL;
      word = strtok(NULL, "+")
  ) {
      words[word_idx++] = word;
  }

I'll add the usual caveats about scanf.

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