C 中 scanf() 的问题

Question

I am learning by myself how to code using the C languange. In order to study this latter a bit more in depth, I'm doing some basic exercises and, due to this, today I have faced a little issue using the scanf() instruction. Infact, the following code:

int main() {

  char inputOne;
  char inputTwo;

  printf("Insert a char: ");
  scanf("%c", &inputOne);
  // &inputOne is the pointer to inputOne.

  printf("Insert another char: ");
  scanf("%c", &inputTwo);

  if (inputOne == inputTwo) {
    printf("You have inserted the same char!\n");
    printf("Chars inserted: %c, %c\n", inputOne, inputTwo);
  } else {
    printf("You have inserted two different chars!\n");
    printf("Chars inserted: %c, %c\n", inputOne, inputTwo);
  }

}

when compiled does not return any error, but when I launch the application on the Terminal, I am not able to insert the second character. Here is an example of what happens:

Macbook-Pro-di-Rodolfo:~ Rodolfo$ /Users/Rodolfo/Documents/GitHub/Fondamenti\ di\ C/esempio-if-else ; exit;
Insert a char: a
Insert a second char: You have inserted two different chars!
Chars inserted: a, 

logout

[Process completed]

Can anybody explain me why this happens?

Answer 1

It takes line feed as input to the second character. You can take the inputTwo again to prevent it:

int main() {

  char inputOne;
  char inputTwo;

  printf("Insert a char: ");
  scanf("%c", &inputOne);
  // &inputOne is the pointer to inputOne.

  printf("Insert another char: ");
  scanf("%c", &inputTwo);
  scanf("%c", &inputTwo);

  if (inputOne == inputTwo) {
    printf("You have inserted the same char!\n");
    printf("Chars inserted: %c, %c\n", inputOne, inputTwo);
  } else {
    printf("You have inserted two different chars!\n");
    printf("Chars inserted: %c, %c\n", inputOne, inputTwo);
  }

}