5

I could not find the answer anywhere else.

%[^\n] - When I run this one, scanf is getting input and terminating after I press enter. ( Probably leaving \n in the input system)

%[^\n]\n - this one is getting the input but scanf is NOT terminating immediately after I press enter like the one above. I hit more enter and it makes more newlines. When I give a character and then press enter it finally terminates. Example: 

int main(void)
{
    char s[100];

    scanf("%[^\n]\n", s);

    printf("%s", s);

    return 0;
}

The results:

enter image description here

Last one:

%[^\n]%*c - When I give some input and press enter. scanf immediately terminates.

How do those 3 work and how do they differ?

Abhishek Bhagate
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Ali Uyanık
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  • aa yes but I would like to know how those 3 actually work, I am wondering the logic behind them. Why they behave in such a way, – Ali Uyanık May 28 '20 at 18:44
  • Possible duplicate: [scanf regex - C](https://stackoverflow.com/q/15664664/643383) – Caleb May 28 '20 at 21:38

1 Answers1

4

All 3 format begin with "%[^\n]".

"%[^\n]" is poor code2 that lacks a width limit and is susceptible to buffer overrun. Use a width limit like "%99[^\n]" with char s[100];.

"%[...]" does not consume leading whitespace like many other specifiers.

This specifier directs reading input until a '\n'1 is encountered. The '\n' is put back into stdin. All other characters are saved in the matching destination's array s.

If no characters were read (not counting the '\n'), the specifier fails and scanf() returns without changing s - rest of the format is not used. No null character appended.

If characters were read, they are saved and a null character is appended to s and scanning continues with the next portion of the format.

"\n" acts just like " ", "\t", "any_white_space_chracter" and reads and tosses 0 or more whitespace characters. It continues to do so until a non-white-space1 is read. That non-whitespace character is put back into stdin.

Given line buffered input, this means a line with non-whitespace following input is needed to see the next non-whitespace and allow scanf() to move on.

With scanf(), a "\n" at the end of a format is bad and caused OP's problem.

"%*c" reads 1 character1 and throws it away - even if it is not a '\n'. This specifier does not contribute to the return count due to the '*'.

A better alternative is to use fgets().

char s[100];
if (fgets(s, sizeof s, stdin)) {
   s[strcspn(s, "\n")] = '\0';  // To lop off potential trailing \n

A lesser alternative is

char s[100] = { 0 };
scanf("%99[^\n]", s);
// With a separate scanf ....
scanf("%*1[\n]");  // read the next character if it is a \n and toss it.

1 ... or end-of-file or rare input error.

2 IMO, worse than gets().

chux - Reinstate Monica
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