Why check if (*argv == NULL)?

Question

In the data structures class that I am currently taking, we have been tasked with writing a web crawler in C++. To give us a head start, the professor provided us with a program to get the source from a given URL and a simple HTML parser to strip the tags out. The main function for this program accepts arguments and so uses argc/argv. The code used to check for the arguments is as follows:

// Process the arguments
if (!strcmp(option, "-h"))
{
    // do stuff...
}
else if (!strcmp(option, ""))
{
    // do stuff...
}
else if (!strcmp(option, "-t"))
{
    // do stuff...
}
else if (!strcmp(option, "-a"))
{
    // do stuff...
}

if ( *argv == NULL )
{
    exit(1);
}

Where "option" has been populated with the switch in argv[1], and argv[2] and higher has the remaining arguments. The first block I understand just fine, if the switch equals the string do whatever based on the switch. I'm wondering what the purpose of the last if block is though.

It could be that my C++ is somewhat rusty, but I seem to recall *argv being equivalent to argv[0], basically meaning it is checking to make sure arguments exist. Except I was under the impression that argv[0] always (at least in most implementations) contained the name of the program being run. It occurs to me that argv[0] could be null if argc is equal to 0, but searching around on Google I couldn't find a single post determining whether or not that is even possible.

And so I turn to you. What exactly is that final if block checking?

EDIT: I've gone with the reasoning provided in the comments of the selected answer, that it may be possible to intentionally cause argv[0] to become NULL, or otherwise become NULL based on an platform-specific implementation of main.

Answer 1

argc will provide you with the number of command line arguments passed. You shouldn't need to check the contents of argv too see if there are not enough arguments.

if (argc <= 1) { // The first arg will be the executable name
   // print usage
}

Answer 2

Remembering just how portable C is, it might not always be running on a standard platform like Windows or Unix. Perhaps it's some micro-code inside your washing machine running on some cheap, hacked environment. As such, it's good practice to make sure a pointer isn't null before dereferencing it, which might have led to the question.

Even so, you're correct. *argv is the same as argv[0], and argv is supposed to be initialized by the environment, if it's provided.

Answer 3

3.6.1/2:

If argc is non-zero those arguments shall be provided in argv[0] though ... and argv[0] shall be the pointer to the initial character of a NTMBS that represents the name used to invoke the program or "". The value of argc shall be nonnegative. The value of argv[argc] shall be 0.

Emphasis mine. argc is only guaranteed non-negative, not non-zero.

This is at entry to main. It's also possible that //do stuff modifies the value of argv, or the contents of the array it points to. It's not entirely unheard of for option-handling code to shift values off argv as it processes them. The test for *argv == null may therefore be testing whether or not there are any command-line arguments left, after the options have been removed or skipped over. You'd have to look at the rest of the code.

Answer 4

just a speculation.

what if your professor is referring to this ??

while(*++argv !=NULL)

    printf("%s\n",*argv);