Thread: Returning zero valued variable from function doesn't work

This is a console program. The user is suppose to type the name of the program at the command line, in this case a.out, followed by the tag -n. The tag -n is to be followed by an arbitrary number of arguments. If you enter two arguments like so:

Code:

a.out -n 1 2

You are congratulated by the program. I you enter 1, 3, 4, 5, 6, 7, 8....etc arguments, you get an error message. There is one problem with this program. It does not work for zero arguments. Typing the following

Code:

a.out -n

should also yield the error message. It does not. I have struggled with this for far too many hours. I come to you now. Do you know why this program doesn't work for zero input arguments???

Code:

#include <stdio.h>

void main(int argc, char *argv[]) 
{

int arg_count, argVal;

	for (arg_count=1 ;arg_count < argc-1; arg_count++)  
	{
		if (!strcmp(argv[arg_count],"-n"))
		{
			counterArgs(argv, arg_count, argc, &argVal);
			if (argVal != 2)
			{	
				fprintf(stderr, "\n\n Error: Either you did not enter two arguments.\n");
				fprintf(stderr, " or you lack basic programming skills.  It appears.\n");
				fprintf(stderr,"that you have entered %d arguments.\n\n", argVal);
				exit(0);
			}
			
			fprintf(stderr,"\n\n Congratulations.  You have entered 2 = %d arguments\n\n", argVal);
		}
	}
}



void counterArgs(char **argv, int arg_count, int argc, int *argVal)
{
*argVal = 0;

	while(*argVal+arg_count < argc - 1 )
	{
		(*argVal)++;
	}	
}

The easiest way i can suggest is to chachk the number of arguments at the begining of program. If argc==2 then print an error message and exit program.

Also, define main as int main() (read FAQ for detaied info), and either put the prototype of your counterArgs() function before main() or move the function definition before main().

It was just a stupid error. I put

Code:

for (arg_count=1 ;arg_count < argc-1; arg_count++)

When I should have put

Code:

for (arg_count=1 ;arg_count < argc; arg_count++)