以下c程序的意外输出

Question

the following code as expected should accept two char values from user. but it just accepts one value of ch1 and then prints "hello".

#include<stdio.h>

int main()
{
    char ch1, ch2;

    printf("Enter a char: ");
    scanf("%c",&ch1);

    printf("Enter second char: ");
    scanf("%c",&ch2);



    printf("Hello");
    return 0;
}

it is not accepting the second value for ch2..what can be the possible reason? As far as i think, it should accept 2 characters.

Answer 1

It just accepts only one char because the the first call to scanf() left a newline in the input stream.

You can ignore it with:

scanf(" %c",&ch2); // note the leading space.

This will ensure the newline from the previous input will be ignored. A white-space in the format string tells scanf() to ignore any number of white-space characters. You might also want to check the return value of scanf() calls in case it failed.

From scanf() :

    .     A sequence of white-space characters (space, tab, newline,
          etc.; see isspace(3)).  This directive matches any amount of
          white space, including none, in the input.