./geany_run_script.sh: Segmentation fault

Question

I want to create a simple program in C that will display a menu, and the user must choose from the choices by choosing a char. This is the code I tried:

#include <stdio.h>

void menu();

int main(int argc, char **argv)
{

    menu();
    return 0;
}

void menu(){
    char choix;
    printf("(C)réer un fichier\n");
    printf("(L)ire un fichier\n");
    printf("(E)crire sur un fichier\n");
    printf("(S)upprimer un fichier\n");
    do{
        choix = tolower(getchar());
        printf(choix);
    }while((choix != 'c') || (choix != 'l') || (choix != 'e') || (choix != 's'));
    printf("end");
}

but when I run my application, I get this message in console:

./geany_run_script.sh: line 5:  6582 Segmentation fault (core dumped) "./main"

And this is a screenshot :

Answer 1

The segmentation fault

printf's argument needs to be a string, but you're passing it a char in

printf(choix);

It's treating that character as a pointer, but it's not going to be a valid pointer, and it's pointing somewhere very low in memory that you should not be trying to access. Instead, you should either pass printf a string with a format directive that expects a char, such as in

printf("%c", choix);

or you could use a function like putchar and do

putchar(choix);

Looping condition problems

However, once you get that worked out, you'll have some problems ending your loop. You end when

(choix != 'c') || (choix != 'l') || (choix != 'e') || (choix != 's')

but every character is not the same as at least one of c, l, e, or s. That is, suppose the user types x. Then the first part (choix != 'c') will be true and you'll quit. The same thing would happen if the user had typed e, which is supposed to be one of your options. If you're trying to loop until the user puts in one of the designated characters, you need to loop while

(choix != 'c') && (choix != 'l') && (choix != 'e') || (choix != 's')

That is, you continue looping as long as it's not c, and it's not l_, and it's not e, and it's not s. You could also phrase this as "loop while it's not ( equal to c or equal tol` or …)", in which case you could loop while

!( choix == 'c' || choix == 'l' || choix == 'e' || choix == 's' )

This mistake actually appears somewhat frequently on Stack Overflow, but it's rather hard to search for. Here's an answered question in Ruby that addresses this issue:

• Test if string is not equal to either of two strings

Answer 2

printf expects a string (char *) as its first argument, you give it a char which it tries to interpret as a pointer. That points to some address it may not read causing a segmentation fault.

In the bottom of your screenshot you can see warnings from your compiler telling you this. Never ignore compiler warnings!

Print your character with putchar or use %c in the printf format string:

printf("%c\n", choix);