LinuxQuestions.org - [C] for looping problem

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[C] for looping problem

I'm almost sure to be right about the following:

If you've got two variables, x and argc, (used for command-line input) and you want to make a loop where x will constantly be raised by one, started with value 1, until the value of argc is reached.
To me, the most neat approach to do that is by using for:

Code:

for ( x = 1 ; x == argc ; x++ )

{



 ( ... )



}

Now I just can't believe my eyes, because on my box, this does not work. All code (the ( ... ) ) simply won't be executed. The only way to get this working, is the following:

Code:

int main(int argc, char **argv)

{

        int x;



        if (argc == 1)

                PrintUsage ();

        else

                for ( x = 1 ; ; x++ )

                {

                        if (x == argc)

                                break;



                        printf ("%i %s\n", x, argv[x]);



                }







        Head ();

        Body ();

        Tail ();

}

:confused: Have I overlooked something?The above example _does_ work, but the first should, too! Right?

Re: [C] for looping problem

This is the miskate you are making

Code:

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

//You used x==argc

{

 ( ... )

}

The for loop should continue until the value of argc has been reached n so the condition should be x<=argc and not x==argc

So:

Code:

for ( x = 1 ; !(x == argc) ; x++)

Should work, too?

Sure, as well as:

Code:

for (x = 1; x != argc; x++)

though, aesthetically, SaTaN's code is the best of the three.

Well, they're not quite the same.

Satan's will execute the loop so long as x is less than or equal to argc, while the other two examples will only execute when x is not equal to argc (meaning that when x is equal to argc, the loop will terminate before it goes through).

True, but that only makes satan's code the better o the three. Because x is incremented by 1 the result in this case is the same.