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#include <stdio.h>
/* The C Programming Language: 2nd Edition
*
* Exercise 1-16: Revise the main routine of the longest-line program so it
* will correctly print the length of arbitrarily long input lines, and as
* much as possible of the text.
*
* Answer: The key to arbitrary limits is buffering. Using a buffer allows you
* to tackle a problem in chunks of memory instead of all at once. It's
* slightly more complicated, but adds usefulness to a program.
*
* This solution follows the spec exactly, by only modifying main(), unlike many
* other solutions on the internet.
*
* Assumptions:
* 1. "arbitrarily long" is interpreted to mean upto the maximum size of an
* integer on the given architechture, e.g. 2^32 unsigned on a 32-bit machine.
* Indeed a string of that size would be larger than 4 gigabytes. It is possible
* to deal with numbers larger than that, but it involves a great deal of work
* abstracting away the concept of an integer, similar to dynamically sized
* arrays.
*
* 2. EOF signal (Ctrl-D) must be given on an empty line
*/
/* demonstrate that we can handle lines much greater than this number*/
#define BUFFSIZE 5
int getlinelen(char line[], int maxline);
void copy(char to[], char from[]);
/* print longest input line */
main() {
/* len of current line, max len seen so far, templen of buffer */
int len, max, templen;
char buffer[BUFFSIZE];
max = len = 0;
while ((templen = getlinelen(buffer, BUFFSIZE)) > 0) {
len += templen;
if (buffer[templen - 1] == '\n') {
if (len > max) {
max = len;
}
len = 0;
}
}
/* The length returned includes the '\n' character. */
printf("\nLen of longest line: %d\n", max);
return 0;
}
/* getlinelen: read a line into s, return length */
int getlinelen(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;
}
/* copy: copy 'from' into 'to'; assume 'to' is big enough */
void copy(char to[], char from[]) {
int i;
i = 0;
while ((to[i] = from[i]) != '\0') {
++i;
}
}
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