Special Case

[ASkoro]

Hey but what about i being sqrt(2)???

[sergiogarcianinja]

I just tried your method, and my compiler is generating a error about a method must return a value, so I fixed it. There is a version without bugs, hope it helps:

if (i < 0)
return 1 - abs(i);
else if (i == 0)
return 1;
else if (i > 0)
return 1 + abs(i);

[sergiogarcianinja]

I a very humble opinion, I think the original developer cared about performance. There is a big and ugly monster living in or closes that will eat us if we write less performing code. The problem is, that almost all developers don't understand about performance and do wrong things. Here, I think he/she are trying to avoid a sum using a comparison. In some cases, like division, it will be a great code.

[YvesDaoust]

Function will return sqrt(2) + 1

[ASkoro]

And for sqrt(-2)????

[YvesDaoust]

SquareRootException + 1

[Lost User]

For the sake of learning here, why do some of the examples use the abs function in their answers. Why not just i++?

[thoiness]

Rewritten:

return (i == 0) ? 1 : i++;

In division, specifically in the denominator, this code eliminates the divide by zero issue. I think the OP (original programmer) had good intentions.

[patbob]

Why not use i++? Obfuscation. The original coder was trying to obfuscate it by using an if statement, so people are running with that theme :)

We can program with only 1's, but if all you've got are zeros, you've got nothing.

[Member 4608898]

And for sqrt(-1/64) do we get indigestion tablets?

[Oscar0]

Bug Alert. I think you meant perhaps: return (i == 0) ? 1 : ++i;

[KP Lee]

ASkoro wrote:

And for sqrt(-2)????

Computers ignore complexity. Or is that irrationality? I'm almost sure that's a complex number. If that is true, what is an irrational number? I know they both exist, but can't definitively define them.

[KP Lee]

Sorry, I fail to see how you fixed it. Computers aren't very good at determining there is an unreachable path. Put an unconditional return 1 - abs(i) + abs(i); after all the if statements should fix it. (Especially if i is uint. Checking for negative numbers is really interesting in that case.)

[KP Lee]

MehGerbil wrote:

Why not just i++?

For one thing that would be the same as returning i. (Unless the i was passed with ref. Then you get two values for the price of one.)

MehGerbil wrote:

For the sake of learning here

That's rich. Trying to learn better coding by studying poor code harder.

[Lost User]

The intent was to write: return i++;

KP Lee wrote:

That's rich. Trying to learn better coding by studying poor code harder.

I think fixing bad code is a great way to learn, especially if you learn the "why" along the way.

[BobJanova]

It should be 'return i + 1'. ++i is a wasteful update of the variable i, assuming it's local (there's some serious issues if it isn't anyway), and i++ is just wrong because it returns i (before the statement) and not i + 1.

[Lost User]

That was interesting. Why is it that ++i is less efficient then returning i + 1? Isn't a calculation made (total of i + 1) made in memory somewhere regardless?

[Vitaly Tomilov]

I'm not surprised, see it all the time. This happens as a result of changing the condition, i.e. the code was written for one condition, then the condition changed, and the code was updated without logical refactoring. Also, some developers like making code temporarily unreachable rather than commenting it out, i.e. putting the code into a block like:

if(false){...code...}

[Albert Holguin]

Developer took special math classes... :-D

[Rotted Frog]

It's likely that compiler optimisations make it irrelevant in this case. But in the general case, she using ++i the value is copied into i after being calculated, then copied into the return variable. Using i+1 only copies it to the return variable (one less copy).