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).

Yep... you should check argc (the number of parameters) before starting your loop...

argc might be 0 and your loop would never catch it.

The easy way, if you need 2 parameters is to check ...

Code:

---

if (argc < 2)
  { printf("Bad command line\n\n");
    exit (1); }

---

Thanks guys. That was the trick. No more compile error. I may be a noob, but I'm still trying to learn how to write clean code that compiles with no errors. :)

Quote:

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Originally Posted by codevyper

Thanks guys. That was the trick. No more compile error. I may be a noob, but I'm still trying to learn how to write clean code that compiles with no errors. :)

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Now comes one of the hardest lessons in programming :) .... "Compiles" does not mean "Works"....

To get past the noob level and to the advanced beginner level, you need to compile code without any warning (with warnings turned on).

Tim S.