Question

I bought a C book called “The C (ANSI C) PROGRAMMING LANGUAGE” to try and teach myself, well C. Anyhow, the book includes a lot of examples and practices to follow across the chapters, which is nice.

Anyhow, the code below is my answer to the books “count the longest line type of program”, the authors are using a for-loop in the function getLine(char s[], int lim). Which allows for a proper display of the string line inside the main() function. However using while won’t work – for a reason that is for me unknown, perhaps someone might shed a light on the situation to what my error is.

EDIT: To summarize the above. printf("%s\n", line); won’t display anything.

Thankful for any help.

#include <stdio.h>
#define MAXLINE 1024

getLine(char s[], int lim) {
    int c, i = 0;

    while((c = getchar()) != EOF && c != '\n' && i < lim) {
        s[++i] = c;
    }

    if(c == '\n' && i != 0) {
        s[++i] = c;
        s[++i] = '\0';
    }    
    return i;
}  

main(void) {
    int max = 0, len;
    char line[MAXLINE], longest[MAXLINE];

    while((len = getLine(line,MAXLINE)) > 0) {
        if(len > max) {
            max = len;
            printf("%s\n", line);
        }
    }
    return 0;
}

Answer 1

You have a number of serious bugs. Here’s the ones I found and how to fix them.

change your code to postincrement i to avoid leaving the first array member uninitialised, and to avoid double printing the final character:

s[++i] = c;
...
s[++i] = c;
s[++i] = '\0';

to

s[i++] = c;
...
// s[++i] = c; see below
...
s[i++] = '\0'; 

and fix your EOF bug:

if(c == '\n' && i != 0) {
    s[++i] = c;
    s[++i] = '\0';
}

to

if(c == '\n')
{
    s[i++] = '\n';
}
s[i] = '\0'

Theory

When writing programs that deal with strings, arrays or other vector-type structures it is vitally important that you check the logic of your program. You should do this by hand, and run a few sample cases through it, providing sample inputs to your program and thinking out what happens.

The cases you need to run through it are:

• a couple general cases
• all the edge cases

In this case, your edge cases are:

• first character ever is EOF
• first character is ‘x’, second character ever is EOF
• first character is ‘\n’, second character is EOF

• first character is ‘x’, second character is ‘\n’, third character is EOF

• a line has equal to lim characters

• a line has one less than lim characters
• a line has one more than lim characters

Sample edge case

first character is ‘x’, second character is ‘\n’, third character is EOF

getLine(line[MAXLINE],MAXLINE])
(s := line[MAXLINE] = '!!!!!!!!!!!!!!!!!!!!!!!!...'
c := undef, i := 0
while(...)
c := 'x'
i := 1
s[1] := 'x' => s == '!x!!!!...' <- first bug found
while(...)
c := '\n'
end of while(...)
if (...)
(c== '\n' (T) && i != 0 (T)) = T
i := i + 1 = 2
s[2] = '\n' => s == '!x\n!!!!'
i := i + 1 = 3
s[3] = '\0' => s == '!x\n\0!!!' <- good, it's terminated
return i = 3
(len = i = 3) > 0) = T (the while eval)
if (...)
len (i = 3) > max = F
max = 3 <- does this make sense?  is '!x\n' a line 3 chars long?  perhaps.  why did we keep the '\n' character? this is likely to be another bug.
printf("%s\n", line) <- oh, we are adding ANOTHER \n character?  it was definitely a bug.
outputs "!x\n\n\0" <- oh, I expected it to print "x\n".  we know why it didn't.
while(...)
getLine(...)
(s := line[MAXLINE] = '!x\n\0!!!!!!!!!!!!!!!!!!!...' ; <- oh, that's fun.
c := undef, i := 0
while(...)
c := EOF
while terminates without executing body
(c == '\n' && i != 0) = F
if body not executed
return i = 0
(len = i = 0 > 0) = F
while terminates
program stops.

So you see this simple process, that can be done in your head or (preferably) on paper, can show you in a matter of minutes whether your program will work or not.

By following through the other edge cases and a couple general cases you will discover the other problems in your program.