Kernighan and Ritchie Exercise 2-2 Debugging? - c

Question

I am working through K&R (2nd edition) for my own edification and encountered the following exercise (exercise 2-2 p42):
Write a loop equivalent to the following without using && or ||:
for (i=0; i<lim-1 && (c=getchar()) != '\n' && c != EOF; ++i)
s[i] = c;
This was my solution:
#include <stdio.h>
/* write a loop equivalent to the following without using && or ||
for (i=0; i<lim-1 && (c=getchar()) != '\n' && c != EOF; ++i)
s[i] = c;
*/
int main()
{
int counter = 0, lim = 1000;
int s[lim], c;
while(counter < lim-1)
{
while((c = getchar()) != '\n')
{
while(c != EOF)
{
s[counter] = c;
}
}
counter++;
}
return 0;
}
I was expecting the indented loops and therefore the entire program to exit normally once it encountered a newline character ('\n') or an EOF character (Ctrl-d on my Linux machine), but to my surprise it happily soldiers on. I tried to debug it using gdb but still could not figure it out.
What am I not seeing?
Addendum: I tried to reverse the sequence of tests the while loops perform and added an if statement to break out of the outer loop if c == '\n' but am still not seeing it! I am also having difficulty trying to run GDB entering text into the command line and simultaneously printing the value of c, even when I tried to link gdb to the pid of a running copy of the executable. I realize that there are probably other ways to solve this exercise, e.g. setting an OK_TO_EXECUTE flag or variable that is true only if all three conditions are met, but I am bothered by the fact that I seem unable to find the bug in a seemingly simple program. This is precisely why I am returning to K&R to go through the book more thoroughly and to solve the exercises properly.
Redone code (still buggy!!!):
#include <stdio.h>
/* write a loop equivalent to the following without using && or ||
for (i=0; i<lim-1 && (c=getchar()) != '\n' && c != EOF; ++i)
s[i] = c;
*/
int main()
{
int counter = 0, lim = 1000;
int s[lim], c;
while((c = getchar()) != EOF)
{
if ( c == '\n')
break;
while(c != '\n')
{
while(counter < lim-1)
{
s[counter] = c;
counter++;
}
}
}
return 0;
}
SOLVED! - I think! I think I have finally figured it out. The inner loops as written in my redone solution will still loop endlessly or at lease till lim is reached. I added break
statements and think I am on my way to a solution.
I am still wrestling with how to run gdb on this problem though; enter the command line entries AND print the value of c. Linking gdb to the pid of the executable still did not work as expected. I even posted a separate question regarding gdb.

Answer 1

but to my surprise it happily soldiers on
You have three nested loops. A newline would terminate one of the inner loops, while the outermost loop would happily carry on (until you've hit Enter lim times).
I can give you a hint: you probably shouldn't be using nested loops for this.

Answer 2

You have added loops that didn't exist in the original ... that's conceptually and logically wrong. The most obvious solution uses break:
for (i = 0; i < lim-1; ++i)
{
c = getchar();
if (c == '\n')
break;
if (c == EOF)
break;
s[i] = c;
}
Or if you're pretending that C doesn't have break, you can do something like this (this is not exactly equivalent because i doesn't have the same value if '\n' or EOF is encountered):
for (i = 0; i < lim-1;)
{
c = getchar();
if (c == '\n')
i = lim-1;
else if (c == EOF)
i = lim-1;
else
s[i++] = c;
}
Or you can use the Pascal approach:
#include <stdbool.h>
...
i = 0;
bool more = i < lim-1;
while (more)
{
c = getchar();
if (c == '\n')
more = false;
else if (c == EOF)
more = false;
else
{
s[i++] = c;
more = i < lim-1;
}
}

Answer 3

With goto
i=0;
loop:
if( i >= lim - 1) goto end;
c = getchar();
if(c == '\n') goto end;
if(c == EOF) goto end;
s[i++] = c;
goto loop;
end:
Without break, goto and with just one for, still without && and ||.
for (i=0; i < lim - 1 ? ((c=getchar()) == '\n' | c == EOF) == 0 : 0; ++i)
s[i] = c;
Update
As noted by #Jim it's way better to set order of execution explicitly by using internal ?:
for (i=0; i >= lim - 1 ? 0 : (c=getchar()) == '\n' ? 0 : c != EOF; ++i)
s[i] = c;