I mistakenly used scanf("%d\n",&val); in one of my programmes, I could not understand the behavior, the function showed.
int main(){
int val;
scanf("%d\n", &val);
printf("%d\n", val);
return 0;
}
Now the program required 2 integer inputs, and prints the first input that was entered.
What difference should that extra \n brings?
I tried to search but couldn't find the answer, even through manual of scanf.
An '\n' - or any whitespace character - in the format string consumes an entire (possibly empty) sequence of whitespace characters in the input. So the scanf only returns when it encounters the next non-whitespace character, or the end of the input stream (e.g. when the input is redirected from a file and its end is reached, or after you closed stdin with Ctrl-D).
From man scanf in my Linux box:
A directive is one of the following:
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.
BTW, %d is also a directive.
So your "%d\n" has two directives, the first one reads the number and the second one... well reads any amount of white space, including your end-of-lines. You have to type any non-white space character to make it stop.
