Date: Mon, 23 Oct 2000 09:58:12 -0700 (PDT)
From: Hannah C. Tang <hctang@cs.washington.edu>
To: cse142-section-bf@cs.washington.edu
Subject: [Section BF] Nested "if"s

Wondering about the "not all control paths return a value" compiler
warning?  Keep reading below ....

Hannah

---------- Forwarded message ----------
Date: Mon, 23 Oct 2000 09:52:56 -0700 (PDT)
From: Hannah C. Tang <hctang@cs.washington.edu>
To: 
Subject: Re: Nested "if"s

In C (and C++) whitespace doesn't matter.  I can write programs that
compile which look like:

 int main(void) {int i = 0; if(i > 0) printf("Hello!\n"); return 0;}

Whitespace just makes programs easier to read:

 int main(void) {
	int i = 0;
	if(i > 0)
		printf("Hello!\n");
	return 0;
 }

The compiler knows what goes into your if-statements only through the
use/non-use of curly-braces, not whitespace.

Sometimes, the compiler will give warnings about "not all control paths
return a value."  What this means is that your if-else/if-else-if's don't
have a final "else":

 if(cond1)
  ...
  return 1;
 else if(cond2)
  ...
  return 2;
 else if(cond3)
  ...
  return 3;
 /* End of function */

You, as the programmer, may know that only one of cond1, cond2, and cond3
may be true, but from the compiler's perspective, it is possible that
neither cond1, cond2, nor cond3 may be true.  So your function could (in
the compiler's point of view) not have a return value!  Thus it flags the
above statements as a warning.  If you draw the control flow diagram, it's
a little clearer what's happening:

(True is left, false is right)

	______cond1______
	|		|
	|	______cond2______
	return 1|		|
	|	return 2	|
	|	|	______cond3______
	|	|	|		|
	|	|	return 3	|
	|	|	|_______________|
	|	|		|
	|	|_______________|
	|		|
	|_______________|
		|
	/* End of function */

Notice that the rightmost branch of your control-path doesn't have a
return value.

If you added a final "else" to your code:

 if(cond1)
  ...
  return 1;
 else if(cond2)
  ...
  return 2;
 else if(cond3)
  ...
  return 3;
 else
  ...
  return 0;
 /* End of function */

Then you would end up with a control-flow diagram which would look like:

(True is left, false is right)

	______cond1______
	|		|
	|	______cond2______
	return 1|		|
	|	return 2	|
	|	|	______cond3______
	|	|	|		|
	|	|	return 3	return 0
	|	|	|_______________|
	|	|		|
	|	|_______________|
	|		|
	|_______________|
		|
	/* End of function */

Notice that all control-flow paths now return a value.

If you're having problems with if-else statements, I'd encourage you to
draw a control-flow diagram; they're a great way to make sure that what
you intend matches up with what you typed.

Hannah

----- Original Message -----
Date: Sat, 21 Oct 2000 21:43:14 -0700 (PDT)
From: 
To: CS 142 TA <hctang@cs.washington.edu>
Subject: Nested "if"s

Hi Hannah,

	How do I know that the compiler will see my nested "if"s
as such? Does it only rely on indentation (along with the use of
braces
'{' and '}'?)