Exercise 5.13 - tail prints the last n lines of input

Question

Write the program tail, which prints the last n lines of its input.

/* Write a Program tail, which prints the last n lines of its input. By default n is 10. let us say; but it can be changed
by an optional argument so that tail -n */

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

#define DEFLINES 10 /* default # of lines to print */
#define LINES   100 /* maximum # of lines to print */
#define MAXLEN  100 /* maximum length of an input line */

void error(char *);
int mgetline(char *,int);

/* print the last n lines of the input */

int main(int argc,char *argv[])
{
    char *p;
    char *buf;  /* pointer to the large buffer */
    char *bufend;   /* end of the large buffer */

    char line[MAXLEN];
    char *lineptr[LINES];   /* pointer to lines read */
    
    int first,i,last,len,n,nlines;

    if( argc == 1)
        n = DEFLINES;

    else if(argc ==2 && (*++argv)[0] == '-')
        n = atoi(argv[0]+1);
    else
        error("Usage: tail [-n]");

    if( n < 1 || n > LINES)
            n = LINES;

    for(i = 0; i < LINES; i++)
            lineptr[i] = NULL;

    if(( p = buf = malloc(LINES * MAXLEN)) == NULL)
        error("tail: cannot allocate buf");
    bufend = buf + LINES + MAXLEN;

    last = 0;
    nlines = 0;

    while((len = mgetline(line,MAXLEN)) > 0)
    {
        if(p+len+1 >= bufend)
            p = buf;
        lineptr[last] = p;

        strcpy(lineptr[last],line);
        if(++last >= LINES)
            last = 0;

        p += len + 1;
        nlines++;
    }

    if( n > nlines)
        n = nlines;

    first = last - n;

    if(first < 0)
        first += LINES;
    
    for(i= first; n-- > 0;i = (i+1) % LINES)
        printf("%s",lineptr[i]);

    return 0;
}

/* error: print error messages and exit */

void error(char *s)
{
    printf("%s\n",s);
    exit(1);
}

/* mgetline: read a line into s and return length */

int mgetline(char s[],int lim)
{
    int c,i;
    
    for(i=0;i<lim-1 && (c=getchar())!=EOF && c!='\n';++i)
        s[i] = c;
    if ( c == '\n')
    {
        s[i] = c;
        ++i;
    }

    s[i] = '\0';
    return i;
}

Run this

Explanation

This program is to print the last n lines of a file, with default being last 10 lines. The program sets aside an array of character pointers (strings) to store the n lines. The LINES value below being the maximum number of lines that can be printed, the default value being 100.

char *lineptr[LINES];   /* pointer to lines read */

The program works by first allocating enough memory for the last n lines in a buffer. Gets each line using mgetline(line, MAXLEN) and then copies each line to an index entry in the lineptr array.

strcpy(lineptr[last],line);

It advances the pointer last at each copy, and when it exceed the maximum count, it just rolls over, starting from 0.

Next, we have to define the logic to print the last n lines. We offset line value appropriately to the number of lines. It is either last - n lines and if that value goes negative, then we increment it to max lines LINES. We start at first value of line and as long as the the line number count represented by n exists, we print the line, decrementing the count at each step.

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