Why does scanf ask twice for input when there's a newline at the end of the format string?

Question

#include 
#include 
#include 

char *method1(void)
{
    static char a[4];
    scanf(\"%s\\n\", a);
    return a;
}

i         


        

Answer 1

From my scanf manual page

White space (such as blanks, tabs, or newlines) in the format string match any amount of white space, including none, in the input. Everything else matches only itself.

Thus with scanf ("%s\n", a) it will scan for a string followed by optional white space. Since after the first newline more whitespace may follow, scanf is not done after the first newline and looks what's next. You will notice that you can enter any number of newlines (or tabs or spaces) and scanf will still wait for more.

However, when you enter the second string, the sequence of whitespace is delimited and scanning stops.

Use scanf ("%s", a) to not scan trailing whitespace.

Answer 2

Don't use the escape sequence in scanf stdio function

     scanf ("%s", a);

Answer 3

you have to remove the \n from the string format of the scanf. It should be

scanf("%s",a);

EDIT: Explanation

the %s means that the scanf reads the input character till it gets a delimiter which should be a white space like space or tab or new line(\n) so the first enter is get as a delimiter for the "%s" and adding the "\n" to the string format "%s\n" means that the scanf will wait 2 newlines the first newline is related to the delimiter of the "%s" and the second newline is related to the\n of the string format.

Answer 4

you can use either of these to avoid the mentioned problem : scanf("%s",a); or scanf("\n%s",a);

Answer 5

use gets() or fgets() instead...alternatively use scanf("%[^\n]s",a);

Answer 6

Remove \n from the scanf format and give an input and it displays the output based on the given output once.