c string array comparison

Question

Working in C, I'm filling an array with char* return values from a function

 char* files[4][12];
 int i = 0;
 for (;;)
    {
      char* file = get_value();
      strcpy(files[i],file);
      i++;
      if (i > 4 || external_condition)
      break;

     }

      // When I break out of
      // my for loop the following
      // code executes

      for (i = 0; i < 5; i++)
       {
        if (files[i] != NULL)
        manipulate(files[i]);
       }

My problem is that if I break out of the first for loop without assigning values to all elements of files, my comparison in the second for loop fails. If only files[0] and files[1] have content, the loop processes files[2],file[3] and files[4] anyway.

One problem is that you are doing strcpy(files[i],file); without allocating any memory for the pointer files[i]. This will result in undefined behavior. — Code-Apprentice
– Code-Apprentice, Commented Sep 1, 2012 at 19:45

rodrigo · Accepted Answer · 2012-09-01 20:19:51Z

3

files is declared as an "array of arrays of pointers to char". Or if you prefer, as a two-dimensional array of pointers to char.

So files[i] is of type "array of pointers to char" but you use it as just a "pointer to char". That is wrong.

That said, it is not clear what you want to do... maybe just:

char files[5][13];

will make more sense. 13 because you likely need 13 char strings (8.3 are 8+3+1=12 plus 1 for the ending NUL), and you seem to use 5 of them. Then initialize them to zero:

memset(files, 0, sizeof(files));

And use the check:

if (files[i][0])

to check if a text is initialized.

edited Sep 1, 2012 at 20:19

answered Sep 1, 2012 at 19:48

rodrigo

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4 Comments

Steve Jessop Over a year ago

Or maybe char files[4][13], considering that an 8.3 filename plus a nul terminator occupies 13 bytes.

rodrigo Over a year ago

@SteveJessop: Good thought. Updated.

obataku Over a year ago

No need for the parentheses for sizeof files.

rodrigo Over a year ago

@veer: I know... sizeof is an operator, but I like it this way, as if it were a function, I find it more readable. You don't need most whitespaces or line feeds, either, but you wouldn't remove them, will you?

obataku · Accepted Answer · 2012-09-01 22:30:42Z

0

char* files[4][12] is a 2D array of char *, not char. Perhaps you meant your code to be as follows... I suggest you listen to what others have said. I'm merely posting a shortened version that still works.

char files[5][12] = { { 0 } };
int i = 0;

do {
  strcpy(files[i], get_value());
} while (++i <= 4 && !external_condition);

while (i) {
  manipulate(files[--i]);
}

edited Sep 1, 2012 at 22:30

answered Sep 1, 2012 at 21:05

obataku

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1 Comment

rodrigo Over a year ago

You have the strcpy() arguments swapped.

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