how i can see if an array with arguments is null in c

Question

i have this code :

  int main(int re, char *args[]){
  int comp=1;
  int j=2;
  int count=0;
 //printf("%d",args[1]);
  while(comp!=0){
       comp=strcmp(args[j],"null");
       j++;
       count++;
  }
}

and i want to see how many string i have in my argument array (args[]), the problem is that a take segmentation fault and i cant find why. When i put NULL instead "null" i get the same result ,segmentation fault. There is a problem with args[j]? or is something else that i dont see? Iam getting out of bounce in my array? i know that the strings begin from args[2] so this is why i put j=2 in the code i put the header file #include to use strcmp

Why are you renaming `argc` to `re`? Also, to check if a pointer is null, use `if (myPointer == NULL) {...}`. And while I'm at it, your `while` loop logic will take you past any eventual arguments, since you only go up and don't stop till you hit 0. — AntonH, Nov 23 '17 at 16:14
`re` is the size of `args`, so use that to check if you've gone out of bounds — Chris Turner, Nov 23 '17 at 16:16
Note that `"null"` is a constant char array containing 5 characters : `'n' 'u' 'l' 'l' '\0'`, and `NULL` is a constant defined in `stdlib.h`. `"null" == NULL` will always be `false` — Hollyol, Nov 23 '17 at 16:25

score 3 · Accepted Answer · answered Nov 23 '17 at 16:32

"null" is a string literal with no special meaning, and NULL is a null-pointer (you could just as well write 0 for it), which is what you mean, but comparing it as a string (with strcmp()) doesn't make any sense. You want to know whether the pointer at args[j] is null (and then, it isn't pointing to anything, so it isn't a string).

Although you get passed the number of arguments in the first parameter to main() anyways, which is almost always simpler to use, the C standard guarantees you that argv[argc] is indeed NULL, so your approach works when implemented correctly. Example:

int main(int argc, char **argv)
{
    int i = 0;
    while (argv[i])  // or more explicit while (argv[i] != NULL)
    {
        puts(argv[i++]); // print this argument
    }
}

Relevant passage in the standard (citing N1570 here) is §5.1.2.2.1 p2:

[...]
-- argv[argc] shall be a null pointer.

josemartindev · Answer 2 · 2017-11-23T16:24:58.223

1

The main function has two parameters,

int main(int argc, char **argv);

argc counts the number of arguments

argv has the arguments as strings in a 2D array.

So, You can just say:

if (argc == 1)
{
   printf("I'm having 0 arguments in Total");
}

The first argc takes up the name of the executable - a.out- . So, if you didn't change the name, a.out is the first argument.

edited Nov 23 '17 at 16:24

answered Nov 23 '17 at 16:16

josemartindev

1,413
9
17

1

You might add that the first entry is the name of the program – Gerhardh Nov 23 '17 at 16:17
1

`argc` will always be at least 1 because `argv[0]` should contain the name of the executable – Chris Turner Nov 23 '17 at 16:17
1

@ChrisTurner is this guaranteed? The C standard only tells `argc` is *non-negative*. I guess you can provoke `argc == 0` on a POSIX system with `execv()`. – Nov 23 '17 at 16:25
1

@ChrisTurner `argc` is not specified to be at least 1. – chux - Reinstate Monica Nov 23 '17 at 16:26
@FelixPalmen Correct me if I'm wrong, but to check the number of arguments, we have to execute the program. Therefore, argc will always be 1 right? – josemartindev Nov 23 '17 at 16:29
1

@joemartin94 C spec also has "If the value of `argc` is greater than zero, the string pointed to by `argv[0]` represents the program name; ..." – chux - Reinstate Monica Nov 23 '17 at 16:30
@joemartin94 *executing the program* isn't automatically the same as passing an argument. – Nov 23 '17 at 16:39

how i can see if an array with arguments is null in c

2 Answers2