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If. Then. Throw. Else. WTF?

java quality

This is the code I could never understand:

if (x < 0) {
  throw new Exception("X can't be negative");
} else {
  System.out.println("X is positive or zero");
}

I have been trying to find a proper metaphor to explain its incorrectness. Today I finally found it.

If-then-else is a forking mechanism of procedural programming. The CPU either goes to the left and then does something or goes to the right and does something else. Imagine yourself driving a car and seeing this sign:

figure

It looks logical, doesn't it? You can go in the left lane if you're not driving a truck. Otherwise you should go in the right lane. Both lanes meet up in a while. No matter which one you choose, you will end up on the same road. This is what this code block does:

if (x < 0) {
  System.out.println("X is negative");
} else {
  System.out.println("X is positive or zero");
}

Now, try to imagine this sign:

figure

It looks very strange to me, and you will never see this sign anywhere simply because a dead end means an end, a full stop, a finish. What is the point of drawing a lane after the dead end sign? There is no point.

This is how a proper sign would look:

figure

This is how a proper code block would look:

if (x < 0) {
  throw new Exception("X can't be negative");
}
System.out.println("X is positive or zero");

The same is true for loops. This is wrong:

for (int x : numbers) {
  if (x < 0) {
    continue;
  } else {
    System.out.println("found positive number");
  }
}

While this is right:

for (int x : numbers) {
  if (x < 0) {
    continue;
  }
  System.out.println("found positive number");
}

There is no road after the dead end! If you draw it, your code looks like this very funny snippet I found a few years ago reviewing sources written by some very well-paid developer in one very serious company:

if (x < 0) {
  throw new Exception("X is negative");
  System.exit(1);
}

Don't do this.