Why is my printf() statement being run twice?

Question

After the initial iteration, the printf() output is shown twice each iteration. Why?

int main()
{
  int ch;
  for( ch = ' '; ch != 'q'; ) {
     printf("Enter a character: ");
     ch = getchar();
  }
  printf("You entered a q!\n");

  return 0;
}

The terminal output is:

Enter a character: w
Enter a character: Enter a character: a
Enter a character: Enter a character: q
You entered a q!

@BLUEPIXY Of course, I know that, but why is the `printf` function in the loop executed apparently twice? — nbro, May 16 '17 at 23:27
@Gilles I retracted my vote to close this question since the other question is asking about what the expected behavior is and how to make it work as the OP wanted. In this case, the OP is also asking why this is happening. — nbro, May 16 '17 at 23:29
@BLUEPIXY So, essentially, we're seeing twice `Enter a character:` in `Enter a character: Enter a character: a` because `\n` is consumed automatically in the iteration after entering "w" as input and pressing enter. This is exactly what I wanted you to tell me and the others. — nbro, May 16 '17 at 23:32

score 4 · Answer 1 · answered May 16 '17 at 23:23

4

You didn't enter w on the command line. You entered w\n. That's two characters.

answered May 16 '17 at 23:23

Bill Lynch

80,138
16
128
173

score 2 · Answer 2 · edited May 16 '17 at 23:25

2

Because getchar reads a character and '\n', not only the character that you typed.

edited May 16 '17 at 23:25

Archmede

1,592
2
20
37

answered May 16 '17 at 23:23

Riccardo Bonafede

610
1
9
17

Manav Dubey · Accepted Answer · 2017-05-16T23:34:18.210

0

As everyone has already stated, getchar() is consuming a newline ('\n') making you have two iterations. A way to fix this is to do this:

int main(){
  int ch;
  for( ch = ' '; ch != 'q'; ) {
     printf("Enter a character: ");
     ch = getchar();
     getchar();
  }
  printf("You entered a q!\n");

  return 0;
}

The reason for the second getchar() is to consume that newline so you won't have a double output of the same thing. Using this method will only work if you are only inputting one character.

edited May 16 '17 at 23:34

answered May 16 '17 at 23:29

Manav Dubey

780
11
26

2

That's not a good way to fix it. It only works if the user entered a single character on the line. A better fix would be to loop over `getchar` until a newline is found. – Gilles 'SO- stop being evil' May 16 '17 at 23:32

score 0 · Answer 4 · answered May 16 '17 at 23:50

As comments and previous answers have stated, getchar() does not read the newline at the end. The quick fix is to add another getchar(); at the end of your loop like this:

int main()
{
  int ch;
  for( ch = ' '; ch != 'q'; ) {
     printf("Enter a character: ");
     ch = getchar();
     getchar();
  }
  printf("You entered a q!\n");

  return 0;
}

If you want a more flexible solution that will work if the user types more than one character, try this:

int main()
{
  int ch;
  for( ch = ' '; ch != 'q'; ) {
     printf("Enter a character: ");
     ch = getfirstchar();
  }
  printf("You entered a q!\n");

  return 0;
}
int getfirstchar() {
  int c = getchar();
  while(getchar() != '\n')
    ;
  return c;
}

I moved reading the character into a separate function so it's more adaptable to future code.

Why is my printf() statement being run twice?

4 Answers4