If i = 0, why is i += i++ 0? | Hacker News Reader

If i = 0, why is i += i++ 0?

stackoverflow.com

8 pointsivoflipse11 comments

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The answer, from the same site:

http://stackoverflow.com/questions/24853/what-is-the-differe...

"++i will increment the value of i, and then return the incremented value."

"i++ will increment the value of i, but return the pre-incremented value."

yep that's 101 stuff eh?

It's a bit more subtle than that.

  i += i++;
  Console.write(i);

could print 0, 1, or 2 depending on the order in which

i++

and

  i += <value>

is evaluated. From the winning answer, what's really happening is:

  int i = 0;
  i = i + i;
  i + 1; // Note that you are discarding the calculation result

This kind of nonsense is the reason why Python doesn't have prefix and postfix increments.

Using gcc or g++ I get "1" not "0" for that expression, as I would intuitively expect. But this is undefined behavior in c/c++: tmp.c: In function ‘main’: tmp.c:3:5: warning: operation on ‘i’ may be undefined [-Wsequence-point]

Eh?

int main() { int i=0; return i += i++; }

Returns 0, as expected.

Why did they close the question?

I still never understand why they close some questions. Quite often I find a very useful answer that's posted on a closed question.

This is for c# not c or c++, fyi

this is correct. i = 0; i+=i++;

(0) + (0)++ = 0 next time around (1) + (1)++ = 2

Unfortunately this is not correct.

If you run

  for (int x = 0; x < 10; x++)
  {
    i += i++;
    Console.WriteLine(i);
  }

i is always 0.

undefined behavior, isn't it? might as well be 42.

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