problem with variable passed as reference

This is completely confusing me. I have a call to a function:

Code:

 case '5': if ((findfree((int)&roomnum)) < 0)
                                                  printf("No available rooms\n");
                                          else
                                                  printf("%d is available\n", roomnum);
                                          break;

and it leaves roomnum with a 10 digit number (when to me) the definition of the function looks like it should give the varaiable roomnum either 1-10 or return -1..

Code:

#include "hotel.h"
#include <unistd.h>
#include <sys/types.h>
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdbool.h>

int fd;
char buf[NLINES];

int
findfree(int room)
{
    ssize_t nread;
    int i, j;
    _Bool vacant;

    if (lseek(fd, (off_t)0, SEEK_SET) < 0)
        perror("problem with lseek in findfree: line 23");

    for (i = 1; i <= MAXROOMS; ++i) {
        if ( (nread = read(fd, buf, sizeof(buf))) < 0)
            perror("problem with read in findfree function: line 21"), exit(-1);

        vacant = true;
        for (j = 0; j < nread; ++j) {
            if (isalpha(buf[j])) {
                vacant = false;
                break;
            }
        }
        if (vacant == true) {
            room = i;
            printf("***%d\n\n", room);
            return 1;
        }
    }

    return -1;
}

Anyone see what's wrong with this?

Your function seems to return 1 on success, -1 on failure. That's it: no room number. It does store a room number to it's local copy (room = i), but this is lost when the function returns. C is a pass by value lanugage... when you say, function(myvar), the function receives the value of myvar in a new variable, not the variable itself.

What your code does is pass not the variable roomnum or its value to the function, but rather the variable's address in memory (type int *, which is incorrectly cast to an int)...

C++ has pass by reference, it looks like this:

Code:

int myfunc(int &var)
{
   var = 2; // Changes to var will appear outside this function.
}

As opposed to:

Code:

int myfunc(int var)
{
  var = 2; // The assignment is only visible to this function.
}

In C, you can use pointers to accomplish the same idea:

Code:

int myfunc(int *var)
{
  *var = 2; // var will be 2 outside the function.
}

Read the FAQ on pointers.