Why %[^\n]s does not work in loop? - c

Question

I was writing a code to take a space-separated string input for a multiple time in between a loop. I saw that %[^\n]s did not work in loop but %[^\n]%*c did. My question is why %[^\n]sdid not work. Here is my code:
#include<stdio.h>
main(){
while(1){
char str[10];
scanf("%[^\n]s",str);
printf("%s\n",str);
}
return 0;
}

Answer 1

The format specifier %[^\n] means "read a string containing any characters until a newline is found". The newline itself is not consumed. When the newline is found, it is left on the input stream for the next conversion.
The format specifier %*c means "read exactly one character and discard it".
So the combination
scanf( "%[^\n]%*c", str );
means "read a string up to the newline character, put the string into the memory that str points to, and then discard the newline".
Given the format %[^\n]s, the s is not part of the conversion specifier. That format says "read characters until a newline is found, and then the next character should be an s". But next character will never be an s, because the next character will always be the newline. So the s in that format serves no purpose.

Answer 2

%[^\n] does not look for space-separated strings. It reads the whole line up until (but not including) the newline.
If your line contains more than 9 characters this causes undefined behaviour by overflowing the buffer.
The buffer overflow could be prevented by writing %9[^\n] but then you have a new issue: on a longer line, %*c would discard the 10th character and the next scan would keep reading from that same line.
Another complicating factor is that if your file contains a blank line, then %[ considers it a matching failure. That means scanf stops, so it does not go on to process %*c. In this case, the newline is not consumed, and the output buffer is not written to at all.
Your code also goes into an infinite loop because you never break out of while(1).
This code shows correct use of ^[:
while (1)
{
str[0] = '\0'; // In case of matching failure
scanf("%9[^\n]", str); // read as much of the line as we can
scanf("%*[^\n]"); // discard the rest of the line
if ( getchar() == EOF ) // discard the newline
break; // exit loop when we finished the input
printf("%s\n", str);
}
If you really do want to read space-separated text then you could use %9s instead of %9[^\n] in my above example. That introduces a new difference though: %s skips over a blank line.
If that's OK then that's OK. If you want to not skip blank lines, then you could use my code above, but add in at the end:
char *p = strchr(str, ' ');
if ( p )
*p = '\0';
Robust string input is difficult in C!