Simple C flow not flowing as expected

Question

I'm teaching myself C. My background is Java. I'm writing a simple console app that is supposed to prompt the user for a number, and then tell the user what number he printed, and then prompt him for another number. The program exits if the user types in 0.

My code is not behaving the way I expect it to. This is the code:

#include <stdio.h>

int main() {
    while (1) {
        printf("Please enter a number: ");
        int this_is_a_number = 0;
        scanf("%d", &this_is_a_number);
        printf("You entered %d\n", this_is_a_number);
        if (this_is_a_number == 0) {
            break;
        }
        getchar();
    }
    return 0;
}

When the program starts, it doesn't prompt me for a number. It does let me type one in. After hitting return, it lets me type in another, and so on and so forth until I return 0. Then it prints out everything that it was supposed to print out earlier, and then exits.

So for example, if I start the program and then type 7, 8, 9, and 0, hitting return after each one, the console should look like this:

Please enter a number:
7
You entered 7
Please enter a number:
8
You entered 8
Please enter a number:
9
You entered 9
Please enter a number:
0
You entered 0

But instead, it looks like this:

7
8
9
0
Please enter a number:
You entered 7
Please enter a number:
You entered 8
Please enter a number:
You entered 9
Please enter a number:
You entered 0

What am I not understanding about flow in C?

On my computer, it works as expected. Maybe your terminal has some problems. — Stefano Sanfilippo, Jan 16 '14 at 21:05
Hmmm. Based on the answers below, I suppose that our different environments must effect different buffer behaviors. — James Dunn, Jan 16 '14 at 21:17
Imagine that, believing that printf actually prints something before it returns. — Eric Lippert, Jan 16 '14 at 23:32

score 4 · Accepted Answer · answered Jan 16 '14 at 21:04

4

This is probably caused by the fact that the output is buffered. Try to flush the buffer after each printf with fflush, eg:

printf(...);
fflush(stdout);
scanf(...);

You can even directly disable buffering through setbuf(stdout, NULL).

answered Jan 16 '14 at 21:04

Jack

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Wow, this worked. I had no idea. So many things I need to think about in C that I've taken for granted in Java! Many thanks. – James Dunn Jan 16 '14 at 21:09

score 3 · Answer 2 · answered Jan 16 '14 at 21:09

You understand the flow control perfectly. You just don't understand what printf and scanf do.

The first problem, that the output from printf is delayed, is because the standard IO library uses buffering. That is, printf doesn't actually perform the expensive operation of writing the data to a terminal until it feels like it, or you tell it to. In this case, fflush is your friend:

printf("Please enter a number: ");
fflush(stdout);

The second problem, that getchar() isn't waiting for your input, is related to scanf. scanf only reads the characters that are required to form its argument -- in this case, an integer. It leaves all of the other characters in the input stream. So, when you type "79" followed by a newline, scanf reads the 7, reads the 9, and leaves the newline in the input stream. Subsequently getchar consumtes that newline, and doesn't wait for anything.

The solution to the second problem is probably just getting rid of the getchar() altogether.

I accidentally left the getchar() in there from a tutorial. Other than what you've said here, I haven't even looked into what it does. Thanks for the insight! — James Dunn, Jan 16 '14 at 21:12

Simple C flow not flowing as expected

2 Answers2