Thread: Compilation warning question

Here's yet another noob question because... well.. I'm a noob at C programming.
I have the following code:

Code:

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

#define MAX_PATH_LEN 128

void readFile(char *name);
char filename[MAX_PATH_LEN];

int main(int argc, const char *argv[])
{
   int argcount;
   for (argcount = 1; argcount < argc; argcount++) {
      if (strcmp(argv[argcount], "-f") == 0) {
         argcount++;
         if (argcount >= argc) {
            printf("ERROR! No filename specified.\n");
            exit(EXIT_FAILURE);
         }
         else {
            strcpy(filename, argv[argcount]);
         }
      }
   }

   printf("My filename: %s\n", filename);
   readFile(&filename);
   exit(EXIT_SUCCESS);
}

void readFile(char *name)
{
   printf("Filename: %s\n", name);
}

If I compile this code it works but I get the following warning:
mytest.c: In function ‘main’:
mytest.c:28: warning: passing argument 1 of ‘readFile’ from incompatible pointer type


If I removed the for loop and hard coded the argv[value] like this:

Code:

int main(int argc, const char *argv[])
{
   strcpy(filename, argv[2]); // Assume it's called with -f and a filename following
   printf("My filename: %s\n", filename);
   readFile(&filename);
   exit(EXIT_SUCCESS);
}

I do not get a compilation warning.

I assume the warning is generated because gcc doesn't see a value for argv[#] at compile time, but am I not coding this right? Is there a way I should be coding it to parse argv that wouldn't generate a warning at compile time? Or is my assumption of what the warning means completely wrong?

Thanks in advance for any help!

filename is an array of char. When passed as an argument, it is converted to a pointer to char, which is what you want. But you passed &filename instead, which is a pointer to an array of MAX_PATH_LEN chars. Hence, just write readFile(filename).