Kernighan 和 Ritchie 练习 2-2 调试？

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, eg 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;