Why won't this simple program compile?

Question

I'm just trying to do a simple test run of executing a program and passing it two parameters.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char* argv[])
{
    start = atoi(argv[1]);
    stop = atoi(argv[2]);
    for(start; start < stop; start++)
    {
        printf("hello");
    }
}

Why is this not working? When I compile it says start and stop is not declared. And when I try to run it as a executable like this ./file 3 5 it says "." is not recognized as an internal or external command. I'm using mingw on my widows cmd prompt. Any ideas why it wont run like I want? That being: compiling, entering executable ./file 3 5 then it running like normal?

Because you haven't given start or stop types. They need to be int start and int stop the first time you assign to them. — aruisdante
– aruisdante, Commented Mar 18, 2014 at 1:35
Do you know what variable declarations are? Because you don't have any. — user2357112
– user2357112, Commented Mar 18, 2014 at 1:35

ciphermagi · Accepted Answer · 2014-03-18 01:34:59Z

3

That's because you didn't declare start or stop. You have to declare variables before you can use them. They don't just show up out of thin air.

answered Mar 18, 2014 at 1:34

ciphermagi

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Comments

Alaeddine · Accepted Answer · 2014-03-18 02:04:12Z

2

try this

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char* argv[])
{
    int start = atoi(argv[1]);
    int stop = atoi(argv[3]);
    for(start; start < stop; start++)
    {
        printf("hello");
    }
}

edited Mar 18, 2014 at 2:04

answered Mar 18, 2014 at 1:57

Alaeddine

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1 Comment

user26347 Over a year ago

argv[2], typo, sorry my mobile is bugging out.

user2404501 · Accepted Answer · 2014-03-18 02:05:05Z

2

The windows shell you're using is a basically DOS-ish thing. ./prog doesn't work because the slash is not a directory separator. You have to use .\prog, or just prog because . is always implicitly in the PATH.

This is in addition to your lack of declaration of start and stop which is well covered by the other answers and comments.

answered Mar 18, 2014 at 2:05

user2404501

Comments

Alberto Coletta · Accepted Answer · 2014-03-18 11:48:57Z

1

Well, you are actually using start and stop variables without declaring them.

You are passing to the main two strings (argv[1] and argv[2]) that are supposed to be integers; as a consequence you need your start and stop variables to be integers as well.

Therefore you have two options (which are the same, just pick your favourite one):

int start = atoi(argv[1]);
int stop = atoi(argv[2]);

or

int start, stop;
start = atoi(argv[1]);
stop = atoi(argv[2]);

answered Mar 18, 2014 at 11:48

Alberto Coletta

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