Exit with char in C

Question

I'm doing a command interpreter and one of the commands must have as result quiting from the program. When user inserts "q", the program should stop executing.

I've written this, but it seems very useless.

Thanks in advance.

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

void exiting (char x){
if (x=='q') exit(0);

int main(){
char x;
scanf("%c", x);
exiting(x);
return 0;
}

Not understanding what the question is. Aside from the missing `}` at the end of `exiting`, there doesn't appear to be any issue. (ETA: and the fact that you should use `&x` instead of `x`) — wolfPack88, Mar 23 '15 at 15:29
Calling `exit(0)` is a [perfectly reasonable](http://stackoverflow.com/questions/2425167/use-of-exit-function) way to halt program execution, but you might want to at least print out to the user that the program is halting because they typed the exit command first. — aruisdante, Mar 23 '15 at 15:31
In the current form, your program will either `exit(0)`or `return 0;`which is the same (for all practical purposes) — DrKoch, Mar 23 '15 at 15:32
@zenith Well, while reading all the comments I applied (mentally) all necessary patches before writing my comment ;) — DrKoch, Mar 23 '15 at 15:36
@wolfPack88 Thank you. I was distracted. Now it compiles and it is working. — cooper, Mar 23 '15 at 15:48
@cooper , I've rolled-back your edits as it made my answer, as well as, many of the comments above meaningless. BTW, Remove `else main()` from your edited code. — Spikatrix, Mar 23 '15 at 16:01
@CoolGuy Thank you very much for your answer. Just tell me... Why should I remove else main() ? It actually seems "not elegant", but I verified that if I do not put that part of the code, any letter can quit the program, isn't that so? — cooper, Mar 23 '15 at 18:12
@cooper , Calling `main()` recursively is a bad idea. The better way would be to use a `do...while` loop. Yes, any letter can quit the program but I am assuming that you will add more code in the program soon. — Spikatrix, Mar 24 '15 at 03:31

Ivo Valchev · Answer 1 · 2017-04-02T20:08:45.630

1

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

void exiting (char x){
    if (x=='q') exit(0);
}

int main(){
    char x;
    scanf("%c", &x); //scanf requires a pointer since it's the only C way to change a value of a variable within a function
    exiting(x);
    return 0;
}

Or, if you want your program to run until you press 'q', that would be the code:

#include <stdio.h>
int main(int argc, char* argv[])
{
    char x;
    do
    {
    scanf("%c", &x);
    }while(x != 'q');
}

edited Apr 02 '17 at 20:08

answered Mar 23 '15 at 18:37

Ivo Valchev

215
3
11

It should be OK now :) – Ivo Valchev Mar 24 '15 at 06:10
Nope. The signature of `main` should be `int main(int argc, char* argv[])` – Spikatrix Mar 24 '15 at 06:27

Spikatrix · Answer 2 · 2015-03-23T16:01:48.590

scanf("%c", x);

Should be

scanf("%c", &x);

Because scanf with %c expects an argument of type char*. You provide the argument x, which is a char and thus, gives the value of x(which is garbage as x isn't initialized) as the second argument instead of giving the address of x for scanf to store the input. This invokes Undefined Behavior. & functions as the address-of operator and gives the address of its operand.

Also, you forgot to add } here:

void exiting (char x){
    if (x=='q') exit(0);
} //this

Exit with char in C

2 Answers2