Ternary and commas

I'm having trouble using '?' in conjunction with ','

Perhaps I'm going about it the wrong way.

Code:

const unsigned long long int stopwatch::read() //if rng, read() will give a current reading, if !rng, it'll return curr.
{
	return rng ?
/*34*/	        (struct timeval now,
		gettimeofday(&now,0),
		timeval & beg = ((timeval*)data)[0],
		now.tv_sec*100000LL+now.tv_usec) - (beg.tv_sec*100000LL+beg.tv_usec))
		   :
/*39*/	        curr;
}

Code:

stopwatch_linux.cpp: In member function ‘const long long unsigned int stopwatch::read()’:
34: error: expected primary-expression before ‘struct’
34: error: expected `)' before ‘struct’
39: error: expected `:' before ‘;’ token
39: error: expected primary-expression before ‘;’ token

Thanks.

I don't think you can create variables on the fly as you just did it. That makes no sense to me. Try declaring it before.

Your parentheses aren't balanced -- one too many ) (don't ask me which one).

You're right. I saw something like this online:

Code:

int a = (int b = 1, ++b);

But I guess that is invalid.

*edit* and thanks, tabstop.

But Desolation's right, no declarations. You can have expressions, but not declarations.

In my opinion this is better written as if/else and just statements, instead of just one long thing.

And surely everyone will be happier if you write this code as:

Code:

const unsigned long long int stopwatch::read() //if rng, read() will give a current reading, if !rng, it'll return curr.
{
	if (!rng) return curr;
        struct timeval now;
	gettimeofday(&now,0),
	timeval & beg = *(timeval*)data;
	return now.tv_sec*100000LL+now.tv_usec) - (beg.tv_sec*100000LL+beg.tv_usec));
}

With this code, you can debug it. And almost anyone can understand what actually happens (in fact, your comment "if rng, read() will ..." now becomes obsolete, as it is clear from the code that is what happens).
--
Mats

Originally Posted by whiteflags

In my opinion this is better written as if/else and just statements, instead of just one long thing.

You and I.

--
Mats

I feel dumb now. I used the ternary to avoid this:

Code:

function
{
   if(a)
   {}
   else
   {}
   return code_never_gets_here;
}

matsp's solution didn't even occur to me. Der. Thanks.

Your fears were unfounded. If you traced that, you will find that return is reached no matter which path is taken.

Well, if you are ever in that situation again, and the pattern in my code doesn't work, then this will:

Code:

   int ret;
   if (a)
      ret = ...
   else
      ret = ...
   return ret;

In some circles, having a single exit from a function is actually the preferred solution.

--
Mats

Thanks, matsp.

Your fears were unfounded. If you traced that, you will find that return is reached no matter which path is taken.

Of course it always returns, but it is inelegant, since it hides the binary behavior of the function. Everything suggested in this thread makes more sense than what I was avoiding.

*edit* Perhaps we misunderstood each other. In my pseudocode, there should have been a 'return' within each bracket set.

Originally Posted by whiteflags

Your fears were unfounded. If you traced that, you will find that return is reached no matter which path is taken.

Yes, and even simple compilers will figure such a simple case out and not say "some paths may not return a value".

--
Mats

Originally Posted by matsp

With this code, you can debug it.

QFE!

Note that the code used is C++, hence the struct keyword in the variable declaration is not necessary.