is this simple program correct? K&R 1-12

This is how I solved the exercise 1-12 from K&R. I have found another solution online, but mine seems a lot easier than the other, even if sometimes it prints too many newlines.
Is this solution acceptable? If not, why? Thank you in advance.

FSX

Code:

#include <stdio.h>

int main(void)
{
	int c;

	while((c = getchar()) != EOF)
	{
		if((c == ' ') || (c == '\t'))
			putchar('\n');
		else
			putchar(c);
	}
	return 0;
}

Originally Posted by fsx

Is this solution acceptable? If not, why?

Since your program is not a correct solution, it is especially not acceptable. Re-read the exercise: "Write a program that prints its input one word per line." Your program occasionally prints empty lines, i.e. lines that don't contain a word (and certainly K&R weren't thinking about the empty word here).

Try to get rid of these empty lines, e.g. by introducing two states "last character was a whitespace" and "last character wasn't". Also have a look at the isspace() function/macro.

mine seems a lot easier than the other, even if sometimes it prints too many newlines.

If you allow your program to occasionally produce incorrect results, here's an even simpler version:

Code:

int main() {}

;-),
Philip

Originally Posted by fsx

This is how I solved the exercise 1-12 from K&R. I have found another solution online, but mine seems a lot easier than the other, even if sometimes it prints too many newlines.
Is this solution acceptable? If not, why? Thank you in advance.

FSX

Code:

#include <stdio.h>

int main(void)
{
	int c;

	while((c = getchar()) != EOF)
	{
		if((c == ' ') || (c == '\t'))
			putchar('\n');
		else
			putchar(c);
	}
	return 0;
}

i executed ur code and it seems to me its working properly.i simply inputted a line and it gave me each word of the line in a new line.

Try inputting two or more lines, Ben. You'll see the code prints additional blank lines.

In any event, fsx, Snafuist is right - printing additional unwanted output hardly qualifies as a correct solution. The objective of writing code is to deliver the required functionality: no more, no less. Having code that is "easier" is not an excuse for delivering slightly different functionality. The programmer is expected to meet requirements, not force the user to accept compromised requirements.

Originally Posted by fsx

Code:

	while((c = getchar()) != EOF)
	{
		if((c == ' ') || (c == '\t'))
			putchar('\n');

What you really want is a "nop" (no-op, no operation) and to continue --

Code:

		if((c == ' ') || (c == '\t'))
			continue;

Except you still need a line break! So maybe

Code:

#include <stdio.h>

int main(void)
{
	int c;
	char flag=0;

	while((c = getchar()) != EOF)
	{
		if((c == ' ') || (c == '\t'))
			if (flag) continue;
			else { putchar('\n');
				flag=1; }
		else {	putchar(c);
			flag=0; }
	}
	return 0;
}

If you are just on chapter 1 you may not know about "continue" (it just starts the loop again without doing anything else).

The use of a flag is a very handy, basic, chapter 1 kind of thing, so if you don't understand what that's about, try to, or ask and I'll explain.

Basically now you could type:

Code:

this       is     a   sentance   with  extra      spaces
Output:
this
is
a
sentence
with
extra
spaces

That was the idea that Snafuist introduced at the beginning. I think you could come up with a better name than "flag" for that variable, though. Maybe something like "saw_whitespace". There's also a logical bug in the program (which was present in my first version too): since the flag is initially zero, there will be a blank line at the beginning of the output if the first character is whitespace. Also, continue isn't really necessary.

Anyway, here's my shot at it.

Code:

#include <stdio.h>
#include <ctype.h>

int main() {
    int c, between_words = 1;

    while((c = getchar()) != EOF) {
        if(isspace(c)) {
            if(!between_words) {
                putchar('\n');
                between_words = 1;
            }
        }
        else {
            putchar(c);
            between_words = 0;
        }
    }

    return 0;
}

Note the use of isspace() from ctype.h. isspace [C++ Reference]

I am still stumbling...

Code:

#include <stdio.h>

int main(void)
{
	int c, pc;

	while((c = getchar()) != EOF)
	{
		if((c == ' ') || (c == '\t') && (pc != c))
			putchar('\n');
		else if((c != ' ') && (c != '\t'))
			putchar(c);
		else
			putchar(c);
		pc = c;
	}
	return 0;
}

What is not working?

I would expect that you will need a nested if-statement:

Code:

   if (c == ' ' || c == '\t')
   {
     if (pc != c)
       putchar('\n');
   }
   else 
       putchar(c);

This will prevent multiple spaces from giving multiple newlines.

--
Mats

Thanks a lot! What I would like to understand is why if I use the && it doesn't works... shouldn't it be similar to another IF statement?

Originally Posted by matsp

What is not working?

I would expect that you will need a nested if-statement:

Code:

   if (c == ' ' || c == '\t')
   {
     if (pc != c)
       putchar('\n');
   }
   else 
       putchar(c);

This will prevent multiple spaces from giving multiple newlines.

--
Mats

Because you do not want the ELSE part to be executed if the second if is false - only if the first if is false.

You could of course add a "if (pc != c)" inside the else-branch instead, but that's less logical.

--
Mats

Originally Posted by dwks

That was the idea that Snafuist introduced at the beginning. I think you could come up with a better name than "flag" for that variable, though.

I called it "flag" because it is a (boolean) flag and I wanted the OP (fsx) to pick up on that (Snafuist refers to this as "two states", which is not quite as specific -- I guess you could see a boolean flag as an implementation of a two-state concept).

You're right about the continue being unnecessary and the initial whitespace issue.

Here's a related question: I usually use a char for boolean values cause it's smaller, but would it be true that an int is actually faster for the processor because it has a four byte boundary? Sort of nit-picking, but this could have a lot of significance in the right place...

Originally Posted by MK27

Here's a related question: I usually use a char for boolean values cause it's smaller, but would it be true that an int is actually faster for the processor because it has a four byte boundary? Sort of nit-picking, but this could have a lot of significance in the right place...

yes it could. you need profiling to prove it or opposite (when the cache size is critical issue)

Originally Posted by matsp

Because you do not want the ELSE part to be executed if the second if is false - only if the first if is false.

You could of course add a "if (pc != c)" inside the else-branch instead, but that's less logical.

--
Mats

Mats got exactly what I was looking for, a simple explanation. I'm new to C and all those pointers, includes and so on are totally alien to my current knowledge. But thanks also for the other answerers.

Originally Posted by MK27

Here's a related question: I usually use a char for boolean values cause it's smaller, but would it be true that an int is actually faster for the processor because it has a four byte boundary? Sort of nit-picking, but this could have a lot of significance in the right place...

Can't bit fields be used for this (making the int size smaller) ? Or is the struct making it larger again?

Code:

#include<stdio.h>

struct foo
{
	int flag:2;
} bar;

int main(void)
{
	bar.flag = 1;
	printf("%d", bar.flag);
	
	return 0;
}

Bitfields won't actually shrink the structure smaller than the actual type. That is to say, even if you set up a bit field of :2, it's still going to be at least an int in size.

Quzah.