using getopt to read till EOF

You can check for 'j' while 'getopt()' is evaluating the arguments, e.g. like

int
main(int argc, char *argv[ ])
{
    extern char *optarg;
    extern int optind, optopt;
 
    bool j_present = false;
 
    while ((c = getopt(argc, argv, "j")) != -1) {
        switch(c) {
        case 'j':
            j_present = true;
            break;
        }
 
    if (!j_present) {
 
      char buf[256];
 
      while(!feof(stdin)) {  // read from stdin till EOF
 
        fgets(buf,sizeof(buf),stdin);
      }
    }
 
    //...
 
    return 0;
}

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Oh, and the same with C++ streams:

#include <iostream>
#include <string>
using namespace std;
 
int
main(int argc, char *argv[ ])
{
    extern char *optarg;
    extern int optind, optopt;
 
    bool j_present = false;
 
    while ((c = getopt(argc, argv, "j")) != -1) {
        switch(c) {
        case 'j':
            j_present = true;
            break;
        }
 
    if (!j_present) {
 
      string buf;
 
      while(!cin.eof()) {  // read from stdin till EOF
 
        getline(cin,buf);
      }
    }

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how do you input an EOF in C?

I think what my question asks is this, if it's just sitting there in the while loop waiting for an EOF from stdin/user, what should the user type in in order for it to go out of that loop?

EOF is defined in stdio.h as

#define EOF     (-1)

so you can just use it that way, e.g.

printf("This text ends with EOF%c", EOF);
 
// or
 
cout << "This text ends with EOF" << EOF;

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Hmm... not sure if that's what I am looking for, when the program enters the while loop below. It just waits there for something to get out of it right? It's waiting for the user to type something, an EOF so it can go out. So to be more specific and a bit dumb. What should I press in my keyboard to get it out of that loop?

  while(!feof(stdin)) {  // read from stdin till EOF
 
        fgets(buf,sizeof(buf),stdin);
      }

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I think we should check if buf has EOF

Hmm..this is kind of an off topic question but somehow related.. say after the -j option it should only take an int, so for example -j 123 , how do I check  if it is an integer and not a char

That you can do like

#include <stdlib.h>
 
bool get_num(char* p,int& num) {
 
  char* pcEnd = NULL,
 
  num = (int) strtol(p,&pcEnd,10);
 
  if (*pcEnd == '\0') return true; // valid number
 
  return false;
}
 
//...
 
    while ((c = getopt(argc, argv, "j")) != -1) {
        switch(c) {
        case 'j':
 
            char* value = (char*) optarg;
            int num;
            j_present = get_num(value,num);
            break;
        }

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 EOF is ctrl-z

hmmm... that's basically for converting a char to an int.. what I want is that if they do something like
-j abasd then I want to print out an error message as the correct way to supply to the option would be -j 1234

Well, if that was '-j abasd' the flag would not be set and they would be asked, but  you could make that

    while ((c = getopt(argc, argv, "j")) != -1) {
        switch(c) {
        case 'j':
 
            char* value = (char*) optarg;
            int num;
 
            if(!get_num(value,num)) {
 
              printf("Error: Needs to be '-j <numeric value>'\n");
            }
            else j_present = true;
 
            break;
        }

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isn't there a simpler approach to this? will get_num store the int value into num?

Yes, it will store the actual value read from '-j' there. And I am afraid it won't get much simpler, since you have to "manually" validate each argument anyway, since "getopt()" cannot do that.

okay.. thanks jkr!