getchar() command used several times

Question

#include <stdio.h>
#include <stdlib.h>
int main(){
    char c1, c2, c3;
    printf("Digite um caractere: \n");
    c1 = getchar();
    printf("Digite um caractere: \n");
    c2 = getchar();
    printf("Digite um caractere: \n");
    c3 = getchar();
    printf("Primeiro caractere: %c\nSegundo caractere: %c\nTerceiro caractere: %c\n", c1);
    system("pause");
    return 0;
}

This code should read 3 characters and using only one printf, show the characters on the screen, however this doing the first reading, I would like to understand why and how to fix. Thanks!

Exercise: Make a program that reads three char characters and then print them one on each line. Use a single printf() command for this.

Answer 1

Your problem happens because you are not familiar with the program input. When you get something like

a
b
c

in fact your program reads as

a\nb\nc

and your program ends up getting

c1 = 'a'
c2 = '\n'
c3 = 'b'

To solve this, you can add getchars that only serve to remove the ' n' from the path by getting the following code:

#include <stdio.h>
#include <stdlib.h>
int main(){
    char c1, c2, c3;
    printf("Digite um caractere: \n");
    c1 = getchar();
    getchar(); //pega o primeiro '\n'
    printf("Digite um caractere: \n");
    c2 = getchar();
    getchar(); //pega o segundo '\n'
    printf("Digite um caractere: \n");
    c3 = getchar();
    printf("Primeiro caractere: %c\nSegundo caractere: %c\nTerceiro caractere: %c\n", c1, c2, c3);
    system("pause");
    return 0;
}

Note: I added more parameters to the printf that were missing

Obs2: Another thing worth mentioning is that if you are using windows, usually the line break happens through two characters, then two getchars would be required for each line break in the input

Answer 2

Thus?

#include <stdio.h>

int main() {
    char c;
    printf("Digite um caractere:\n");
    char c1 = getchar();
    while ((c = getchar()) != '\n' && c != EOF) { }
    printf("Digite um caractere:\n");
    char c2 = getchar();
    while ((c = getchar()) != '\n' && c != EOF) { }
    printf("Digite um caractere:\n");
    char c3 = getchar();
    while ((c = getchar()) != '\n' && c != EOF) { }
    printf("Primeiro caractere: %c\nSegundo caractere: %c\nTerceiro caractere: %c\n", c1, c2, c3);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Just failed to print the other two, it is not enough to have intention to show you have to pass the variable with the value for the print to be made.

I simplified a little too.

In fact the getchar() does not work the way you expect it to. In fact every C data entry mechanism is confused by the way it manages the buffer. And there are many people who try magical solutions. Some work only in certain situations but as she sees it working she thinks it’s right. It’s full on the internet showing wrong. So it’s best to do the simple or make a function of yours that always works the way you expect.

I did a cleaning of buffer in the hand that always works, even if it is ugly.

Anyway, I don’t know if this is the purpose of the exercise. Is he saying that he needs to read the 3 characters separately? No, then you can do it another way. It is only specific in how it should print. So it works perfectly:

#include <stdio.h>

int main() {
    printf("Digite três caracteres:\n");
    char chars[4];
    scanf("%3s", chars);
    printf("Primeiro caractere: %c\nSegundo caractere: %c\nTerceiro caractere: %c\n", chars[0], chars[1], chars[2]);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

You can read more on How to read from stdin in C?. And how to use the getchar() in the right way (yes, it’s confusing).