Passing char * into fopen with C

I'm writing a program that passes data from a file into an array, but I'm having trouble with fopen (). It seems to work fine when I hardcode the file path into the parameters (eg fopen ("data/1.dat", "r");) but when I pass it as a pointer, it returns NULL.

Note that line 142 will print "data/1.dat" if entered from command line so parse_args () appears to be working.

132 int
133 main(int argc, char **argv)
134 {
135   FILE *in_file;
136   int *nextItem = (int *) malloc (sizeof (int));
137   set_t *dictionary;

138
139   /* Parse Arguments */
140   clo_t *iopts = parse_args(argc, argv);
141
142   printf ("INPUT FILE: %s.\n", iopts->input_file); /* This prints correct path */
143   /* Initialise dictionary */
144     dictionary = set_create (SET_INITAL_SIZE);
145
146   /* Use fscanf to read all data values into new set_t */
147   if ((in_file = fopen (iopts->input_file, "r")) == NULL)

148   {
149     printf ("File not found...\n");
150     return 0;
151   }

Thanks!
Rhys

MORE: If I try to print the string after I run set_create() (ln 144), the string doesn't print. (But there isn't any reference to the string in the function at all...)

 47 set_t *
 48 set_create(int size)
 49 {
 50   set_t *set;
 51
 52   /* set set_t members */
 53   set->items = 0;
 54   set->n_max = size;
 55   set->lock = FALSE;

 56
 57   /* allocate memory for dictionary input */
 58   set->data = (int *) malloc (size * sizeof (int));
 59
 60   return set;
 61 }

It does work if I call this function after fopen ().
I can't see how this is affecting the filename though...

Thanks again.

Answer

Your new code shows that you are writing to invalid memory. set is a pointer but you never initialize it. You're overwriting some random memory and thereby destroying the pointer to the string that you're passing to fopen().