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I've written a simple implementation of tail as part of reading The C Programming Language by Kernighan & Ritchie.

The question states to write tail, which prints the last n lines of its input. By default, n is 10 but should be able to be specified by writing: tail -n to print the last n lines. I don't think the program has to open and read files, as that hasn't been covered in the book as of yet.

I chose to take input as tail -n <number> and wrote the code for parsing the arguments in (hopefully) that would allow for addition of more arguments at a later date easier using the switch statement. However right now, if there were more arguments (say x and z) one could write tail -xzn 10 which would be valid, printing the last 10 lines of input - which maybe could be viewed as a problem.

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

#define MAXLEN 1000

int tail(char *buffer[], int n);
int read_line(char *line, int max);
void print_lines(char* buffer[], int n);

static int lines_read = 0;

int main(int argc, char* argv[]) {
  int number_of_lines = 10;
  char c;

  while(--argc > 0 && (*++argv)[0] == '-') {
    while((c = *++argv[0])) {
      switch(c) {
        case 'n':
          if(argc-1 > 0)
            number_of_lines = atoi(*(argv+1));
          break;
      }
    }
  }  

  char *line_buffer[number_of_lines];

  if(tail(line_buffer, number_of_lines) < 0) {
    printf("ERROR: failed to allocate memory for a line.\n");
    return -1;
  }

  printf("Output:\n"); 
  print_lines(line_buffer, number_of_lines);

  return 0;
}

int tail(char *buffer[], int n) {
  int characters_read = 0;
  char *p, line[MAXLEN];

  while((characters_read = read_line(line, MAXLEN)) > 0) {
    if((p = malloc(sizeof(char)*characters_read)) == NULL) {
      return -1;
    }
    strcpy(p, line);
    buffer[lines_read++ % n] = p;
  }

  return 0;
}

int read_line(char *line, int max) {
  int c, chars_read = 0;

  while((c = getchar()) != EOF && c != '\n' &&  ++chars_read < max-1)
    *line++ = c;

  *line = '\0';

  return chars_read;
}

void print_lines(char *buffer[], int n) {
  for(int i=0; i < (lines_read > n ? n : lines_read); i++)
    printf("%s\n", buffer[lines_read > n ? lines_read++ % n : i]);
}
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  • \$\begingroup\$ <limits.h> has an implementation-defined LINE_MAX that you can piggyback off of instead of defining your own MAXLEN. \$\endgroup\$
    – Andrew
    Commented Jan 25, 2018 at 19:47
  • \$\begingroup\$ @AndrewPiliser: Do you have a reference for <limits.h> defining LINE_MAX? I can't seem to find it in my copy of the standard. \$\endgroup\$ Commented Jan 25, 2018 at 20:06
  • \$\begingroup\$ @JerryCoffin That might be a Linux-specific thing, I found it here: pubs.opengroup.org/onlinepubs/7908799/xsh/limits.h.html. \$\endgroup\$
    – Andrew
    Commented Jan 25, 2018 at 22:12

1 Answer 1

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\$\begingroup\$
  • Avoid globals. lines_read is naturally suited to return lines_read.

  • tail leaks memory. You shall free the line pointed by buffer[....] before reassigning it.

  • sizeof(char) is guaranteed to be 1.

I don't remember at which point K&R offer this exercise. In any case,

  • The line-reading loop is better expressed as fgets.

  • malloc/strcpy combination is a long way to say strdup.

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  • \$\begingroup\$ Is there a reason to use fgets instead of getline? \$\endgroup\$
    – Andrew
    Commented Jan 25, 2018 at 19:46
  • \$\begingroup\$ @AndrewPiliser: all the usual ones, such as it's already well documented, and tested, and everybody who knows C immediately recognizes what it is and what it does. There is another minor detail: his getline looks like if a line is too long, it'll read a single line as multiple lines, and give no warning that it's doing so. fgets can read partial lines, but gives you a warning when it does so (the buffer you read will have a new-line if and only if you read the entire line). \$\endgroup\$ Commented Jan 25, 2018 at 20:17
  • \$\begingroup\$ @AndrewPiliser I was not even sure that fgets is within the scope of this exercise. get line is definitely out. \$\endgroup\$
    – vnp
    Commented Jan 25, 2018 at 20:46

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