Where is scanf() reading input from, if not from the keyboard?

Question

I'm a novice programmer getting introduced to C and I'm missing something fundamental about the way my scanf() works. I want to read a single int from the keyboard with code like this:

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

int main(void) {
    int userBookSelection;
    scanf("%i", &userBookSelection);
    printf("Printing userBookSelection: %i", userBookSelection);

    return EXIT_SUCCESS;
}

When I run the code, the console stays black until I stop debugging. There is never a cursor waiting for keyboard input. When I stop debug I can see this output in the console, same every time:

Printing userBookSelection: 2130567168

I'm debugging in Eclipse with MinGW GCC compiler on Windows. The code syntax seems to be correct -- is it possible there's something wrong in my build path to make this happen? I need to know why scanf() isn't reading for keyboard input.

The program as shown is correct ISO C (up to a point: scanf is broken-as-specified, and should never be used in production -- but for this kind of learning exercise it's fine) and should work as intended. Exactly what compiler, debugger, console, operating system, etc. are you using? Leave nothing out. — zwol

J.M. J.M. · Accepted Answer · 2017-09-08T17:38:05

So I've gotten a line of code from my professor which takes care of this bug -- whether it's a necessary solution particular to Eclipse and/or MinGW I'm not sure. In any case, here's the code with the additional line:

int main(void) {
    int userBookSelection;
    setvbuf (stdout, NULL, _IONBF, 0);//<---The magic line

    scanf("%i", &userBookSelection);
    printf("Printing userBookSelection: %i", userBookSelection);

    return EXIT_SUCCESS;
}

I'd appreciate any additional wisdom on what's going on, what setvbuf() is doing and how scanf() works more fundamentally.

Where is scanf() reading input from, if not from the keyboard?

1 Answers