在一行中读取 3 个字符，直到输入特定字符

Question

I'm trying to use scanf to read 3 inputs in a line until it encounters a character 'E' in a newline. Why it doesn't stop scanning until I type in an another character after the character 'E' ?

char s[200];

char ch='A';
int ind=0;

while(ch!='E')
{
    scanf("%c ",&ch);
    s[ind]=ch;
    ind=ind+1;

}
printf("%c",s[2]);

My result image

Answer 1

It's because you have a trailing space in the scanf format string.

That will lead scanf to read and ignore all white-space (spaces, tab, newlines) until it reaches a non-space character.

One simple solution is to use leading space in the format string:

scanf(" %c",&ch);
//     ^
// Note leading space

I haven't added any error checking, but you should really check what scanf returns . I also recommend you add it as a part of the loop condition itself:

char ch;

while (scanf(" %c", &ch) == 1 && ch != 'E')
{
    // Use ch
}

---

Another possible solution is to use a character-reading function like fgetc or getc or getchar . But do note that these return an int which is important because you also need to remember to check the character returned against EOF :

int ch;  // Need to be an int for the EOF check to work

while ((ch = getc(stdin)) != EOF && ch != 'E')
{
    if (isspace(ch))
        continue;  // Don't bother with any kind of space

    // Use ch
}

While using getc as shown above might seem to be more work, it's also more flexible and give you greater control over how to handle different character and character classes.

Answer 2

You can use getc or fgetc for better answer. Here I am pasting a working code.

#include <stdio.h>

int main ()
{
  char s[200];

  char ch = 'A';
  int ind = 0;

  while (ch != 'E')
    {
      ch = getc(stdin); //Instead of scanf you can use getc
      s[ind] = ch;
      ind = ind + 1;

    }
  printf ("%c", s[2]);
  return 0;
}