Argc/Argv C Problems

Question

Hey all,

If I have the following code:

main(int argc, char *argv[]){
 char serveradd[20];
 strcpy(serveradd, argv[1]);
 int port = atoi(argv[2]);
 printf("%s %d \n", serveradd, port);

The first two arguments to the command line are printed. However, if I do this:

 char serveradd[20];
 strcpy(serveradd, argv[1]);
 int port = atoi(argv[2]);
 char versionnum[1]; 
 strcpy(versionnum, argv[3]);
 printf("%s %d %s \n", serveradd, port, versionnum);`

The first argument (serveradd) does not print out to the screen and is not being stored... Why is this happening and how can I fix it? Thanks!

Answer 1

char versionnum[1];  
strcpy(versionnum, argv[3]); 

Wild guess but you are smashing the stack with these lines. Make versionnum bigger; as it stands, it can only safely hold the empty string.

Answer 2

Using strcpy without testing the length using strlen of the string you are copying to be sure it will fit in the destination buffer is a very bad idea. You should use strncpy, which is length-checked and will not overrun the destination buffer.

You cannot store a string in a char array of length one, because the null terminator takes one character, so all you could store would be the null character. It is all but guaranteed that you are overrunning your buffer.

Answer 3

You're probably nuking memory with

char versionnum[1]; 
strcpy(versionnum, argv[3]);

Given that any non-empty null-terminated string is going to be > 1 character long.

Never use straight strcpy(); use strncpy() instead (note the - 1 to preserve space for the null terminator):

char serveradd[20] = { 0 };
strncpy(serveradd, argv[1], sizeof(serveradd) - 1);
int port = atoi(argv[2]);
char versionnum[2] = { 0 }; 
strcpy(versionnum, argv[3], sizeof(versionnum) - 1);
printf("%s %d %s \n", serveradd, port, versionnum);

Answer 4

It works for me in the way you described. My guess is that you're over the character limit in one of the strings. I'd use strncpy instead of strcpy.

Answer 5

I think the answers above resolved your problem, but in the future try being a little more generous with memory allocation when you are debugging. If you think you need 5 elements, create a define make it 10, and later change it to 5 or 6 if you want to have room for the null terminator. Also if you ever find yourself allocating an array of length one, stop and ask why. If you have one element, just use a plain variable.

Whenever I have problems with argc and argv the first think I do is print them out to see if I have what I think should.

Answer 6

Try strdup()

An alternative is to use strdup() which allocates memory on the heap so that your can't overflow the buffers. In your case, since the variables will exist for the life of the program anyway, this is not so bad, or you can use free() to deallocate the memory right after the last time you use the variables. Make sure you don't use the variables after that! Below is some sample code.

---

#include <stdio.h>
#include <stdlib.h>
#include <strings.h>

int main(int argc, char *argv[])
{
      char *serveradd  = NULL;
      int   port;
      char *versionnum = NULL;

      if (argc != 4)
      {
            printf("Usage: %s SERVERADD PORT VERSIONNUM\n", argv[0]);
            exit(126);
      }

      serveradd  = strdup(argv[1]);
      port       = atoi(argv[2]);
      versionnum = strdup(argv[3]);

      if ((serveradd == NULL) || (versionnum == NULL))
      {
            perror("Could not allocate command line arguments");
            exit(125);
      }

      printf("%s %d %s \n", serveradd, port, versionnum);

      free(serveradd);
      free(versionnum);
      serveradd  = NULL;
      versionnum = NULL;
      exit(0);
}