Why use 4096 elements for a char array buffer?

Question

I found a program that takes in standard input

int main(int argc, char **argv) {
    if (argc != 2) {
        fprintf(stderr, "Usage: %s <PATTERN>\n", argv[0]);
        return 2;
    }

    /* we're not going to worry about long lines */
    char buf[4096]; // 4kibi

    while (!feof(stdin) && !ferror(stdin)) { // when given a file through input redirection, file becomes stdin
        if (!fgets(buf, sizeof(buf), stdin)) { // puts reads sizeof(buf) characters from stdin and puts it into buf; fgets() stops reading when the newline is read
            break;
        }
        if (rgrep_matches(buf, argv[1])) {
            fputs(buf, stdout); // writes the string into stdout
            fflush(stdout);
        }
    }

    if (ferror(stdin)) {
        perror(argv[0]); // interprets error
        return 1;
    }

    return 0;
}

Why is the buf set to 4096 elements? Is it because the maximum number of characters on each line can only be 4096?

Answer 1

The answer is in the code you pasted:

/* we're not going to worry about long lines */
char buf[4096]; // 4kibi

Lines longer than 4096 characters can occur, but the author didn't deem them worth caring about.

Note also the definition of fgets:

fgets() reads in at most one less than size characters from stream and stores them into the buffer pointed to by s. Reading stops after an EOF or a newline. If a newline is read, it is stored into the buffer. A terminating null byte (\0) is stored after the last character in the buffer.

So if there is a line longer than 4095 characters (since the 4096'th is reserved for the null byte), it will be split across multiple iterations of the while loop.