%[^\n] in scanf means for C language?

Thank you to Mr. Rohit Adalinge

For asking this question.

#include <stdio.h>

#include <stdlib.h>

int main() {

    char line[100];

    scanf("%[^\n]",line);

    printf("Hello,World\n");

    printf("%s",line);

    return 0;

}

#include <stdio.h>

#include <stdlib.h>

//Case 1

/*

int main() {

    char line[100];

    scanf("%[^\n]",line);

    printf("Hello,World\n");

    printf("%s",line);

    return 0;

}

*/

/*

Output:

C:\TURBOC3\BIN>TC

SVERI Pandharpur

Hello,World

SVERI Pandharpur

*/

//Case 2

// with %s format specifier it stores only first word

/*It means after pressing of spacebar key whichever word will be there

will not be considered.*/

int main() {

    char line[100];

    clrscr();

    scanf("%s",line);

    printf("Hello,World\n");

    printf("%s",line);

    return 0;

}

/*

Output:

SVERI Pandharpur

Hello,World

SVERI

*/

Note:

In such cases instead of using %[^\n] in scanf, we can use gets() function which will work similar to %[^\n] in scanf ().

In short:

The scanf("%[^\n]", line) has the specifier "%[^\n]". 

It scans for unlimited number of characters that match the scan-set ^\n. 

If none are read, the specifier fails and scanf() returns with line unaltered. 

If at least one is read, all matching are read and saved. 

A null character is appended.

For more details please refer:

https://stackoverflow.com

Question was interesting. Great learning for me though this question.

#sdbhosale