CODEWITHSUNDEEP
c
eastboundr
eastboundr

Reputation: 1877

Why do I get a floating point exception?

While trying to take some arguments for C. I found it really difficult to get argv[] to work. I have:

int main(int argc, char *argv[])
{
  void updateNext();
  void fcfs();
  void spn();
  void srt();

  fp = fopen(argv[0],"r");
  op = fopen("output.dat","a+");

  if (strcmp(argv[1],"FCFS")!=0)
  {
    fcfs();
  }

  if (strcmp(argv[1],"SPN")!=0)
  {
    spn();
  }

  if (strcmp(argv[1],"SRT")!=0)
  {
    srt();
  }
}

I would like to enter something in a format of myprog input.data FCFS, but the above code gives me an error for "float point exception" the exception is gone after I hard code input.dat as a string in the program. Something wrong with argv[0] perhaps?

Upvotes: 0

Views: 431

Answers (2)

Greg Hewgill
Greg Hewgill

Reputation: 994599

In C, argv[0] is the name of your program (or more precisely, the first word the user typed on the command line to run your program, if run from a shell).

So, avoiding argv[0] for your purposes, you'll want to look at argv[1] for the file name and argv[2] for the other parameter.

This would have been clear if you had used a debugger to trace through your program, or simply printed the value before you used it:

printf("using file name %s\n", argv[0]);
fp = fopen(argv[0],"r");

It's also a good idea to check that you have sufficient command line parameters by validating argc before accessing argv:

if (argc < 3) {
    fprintf(stderr, "not enough command line parameters\n");
    exit(1);
}

Upvotes: 7

cnicutar
cnicutar

Reputation: 182734

In C argv[0] is typically the name with which the program was called. You're looking for argv[1] and argv[2].

As side notes:

  • you should be checking argc before touching argv
  • there's no guarantee argv[0] contains the name of the program or even something sensible

Upvotes: 5

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