Question

printf is ouput is being destyoryed by a “\n” after I run scanf. I am attempting to use.

scanf ("%*c");

to chomp the “\n” but its not working… here is the code

    printf("Enter char float int char:", char4, deci2, num2, char5);
    scanf ("%c %f %d %c", &char4, &deci2, &num2, &char5);
    scanf ("%*c");
    printf("You entered: '%c' %.3f %d '%c' " ,char4 ,deci2, num2, char5 );

and it outputs to

    Enter char int char float:a 5 a 5.5
    You entered: 'a' 5 'a' 5.500 
    Enter char float int char:a 5.5 6 b
    You entered: '
    ' 0.000 0 ''

Answer 1

The line

scanf ("%c %f %d %c", &char4, &deci2, &num2, &char5);

is picking up the stray newline left in the input buffer from your previous scanf call. You can work around that by putting a space in front of the first %c:

scanf (" %c %f %d %c", &char4, &deci2, &num2, &char5);

This will tell scanf to skip over any leading whitespace (blanks, newlines, tabs, etc.) before reading the next non-whitespace character.

The line

printf("Enter char float int char:", char4, deci2, num2, char5);

is a bit of a head-scratcher; it won’t cause any problems (the excess arguments are evaluated, but otherwise ignored), but it looks wrong, and indicates some confusion.