%[^\n]s in scanf does not wait for input and gets skipped

Question

In the loop in the code below, scanf("%[^\n]s",array) is not working. It does not wait for input and gets skipped. But a space before % fixes the issue. Why?

Here is the wrong program:

#include <string.h>
#include <stdio.h>

int main()    
{    
    int t;
    scanf("%d",&t);
    while(t--){
        char arr[199];
        scanf("%[^\n]s",arr);
        printf("%s",arr);
    }
    return 0;
}

Here is right code:

#include <string.h>
#include <stdio.h>

int main()
{    
    int t;
    scanf("%d",&t);
    while(t--){
        char arr[199];
        scanf(" %[^\n]s",arr);
        printf("%s",arr);
    }
    return 0;
}

Why does it need a space before % for it to work as expected?

Answer 1

First of all, the trailing s is not part of the %[ format specifier, so remove it and lets talk about %[^\n].

Now, what %[^\n] tells scanf to do is scan everything until a newline character ('\n') or EOF, whichever comes first, and stores it in its corresponding argument, in this case, arr.

And here is the catch: %[^\n] fails if the first character to be read is a \n.

'Wait', you say. 'I did not type in a lone enter. So, why'd it fail?'. True. You did not. But remember the Enter you pressed for the previous line? Turns out, the previous call to scanf grabs everything until the \n, leaves the \n there and returns. So, in the next iteration of the loop, the scanf sees this \n character left over by the previous call to scanf, fails and returns 0.

As for the space, it is a whitespace character. And a whitespace character in scanf instructs it to scan and discard all whitespace characters until the first non-whitespace character. So, it removes the \n (since it is a whitespace character) and scanf will wait for further input.