./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 to l` 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);