K&R solution?

[deadhippo]

I was doing one of the K&R exercises which is to write a loop equivalent for the following short program without using || or &&. Usually my solutions are pretty similar to the solution book answer or some of the solutions I can find on the web but this time it is very different. Am I way off with this solution?

for (i = 0; i < lim -1 && (c = getchar()) != '\n' && c != EOF; ++i)
		s[i] = c;

to

        i = 0;
	while (i < lim -1)
		if (c = getchar() != '\n')
			if (c != EOF)
				++i;
        s[i] = c;

[matsp]

I'm assuming yours is the second one, in which case you are missing the condition to break the loop when you get a end-of-line or end-of-file.

--
Mats

[root4]

I guess the 1st solution is the correct one and the second is your attempt?
Clearly they don't do the same thing. The first one fill the array with characters
read from the input stream until the array is full or there's nothing more to read
or the last character was a new line. The second solution fill the array and
everything should be fine until the EOF or '\n'... in which case, if the array is not
full yet, we should fall into an infinite loop as the index is no longer incremented
and the current item in the array is overwritten by EOF or '\n'.
The particular case in which the last character read is also the last index of the
array should work fine however...

[deadhippo]

Thanks. Back to the drawing board.

[Elysia]

if ( (c = getchar()) != '\n' )

Just a note, but it's better to do it like this. Otherwise you would get a warning in Visual Studio saying assignment within if. And we'd like to write code as clean as possible and as portable as possible that works without warnings in as many compilers as possible.

Just thought I'd mention. Carry on!

[deadhippo]

Ok, I have another idea.

	i = 0;
	loop = 1;
	while (loop == 1)
		if (i >= lim-1)
			loop = 0;
		else if ((c = getchar()) == '\n')
			loop = 0;
		else if (c == EOF)
			loop = 0;
		else
		{
			s[i] = c;
			++i;
		}

I think this might work but it is very similar to the answer in Tondo and Gimpel.

Thanks Elysia, I had been reading that that was good practice too.

[matsp]

if ( (c = getchar()) != '\n' )

Just a note, but it's better to do it like this. Otherwise you would get a warning in Visual Studio saying assignment within if. And we'd like to write code as clean as possible and as portable as possible that works without warnings in as many compilers as possible.

Just thought I'd mention. Carry on!

Not to mention that the assignment does the wrong thing, as c will be equal to (getchar() != '\n'), because the comparison operator has higher priority than the assignment.

--
Mats

[matsp]

Ok, I have another idea.

	i = 0;
	loop = 1;
	while (loop == 1)
		if (i >= lim-1)
			loop = 0;
		else if ((c = getchar()) == '\n')
			loop = 0;
		else if (c == EOF)
			loop = 0;
		else
		{
			s[i] = c;
			++i;
		}

I think this might work but it is very similar to the answer in Tondo and Gimpel.

Thanks Elysia, I had been reading that that was good practice too.

I can't see anything wrong, but I would simplify the if/else tree a bit, and use a different condition:

        i = 0;
	while (i < lim-1)
        {
		if ((c = getchar()) == '\n' || c == EOF)
			break;
		s[i] = c;
		++i;
        }

--
Mats

[deadhippo]

Thanks Mats, but according to the exercise we can't use || or &&.

[matsp]

Thanks Mats, but according to the exercise we can't use || or &&.

Ok, but you can reduce the number of if/else branches to two by just splitting that "|| c == EOF" into it's own if-statement.

--
Mats

[deadhippo]

Thanks. I'll try that.

[SirNot]

why even bother splitting it up into if/elseif cases. just use the fact that conditionals such as == and != represent a boolean value.

ie. something like

while((cond1) + (cond2) == 2) ...

[matsp]

why even bother splitting it up into if/elseif cases. just use the fact that conditionals such as == and != represent a boolean value.

ie. something like

while((cond1) + (cond2) == 2) ...

That is just bypassing the rule of not using || and &&, isn't it? I think the rule was there to show how much simpler [and sometimes more complex] the code becomes when you use or don't use those - not "find another way to make the same thing", even if your suggestion is quite creative.

--
Mats