Poser update

A few updates on the results of my “poser” post.

First, anybody who said “anybody who coded that deserves whatever they get” is absolutely correct, but gets no points because that wasn't the answer we were looking for.

I thought it was an interesting question for two reasons (most of which Wes covered in his response, but you may not have read it).

First, when you write:

x += 5;

the compiler write this as

x = x + 5;

which obviously changes the order of evaluation, and the answer to the question.

The second is that C# evaluates from left to right, so you can reason about how expressions are evaluated.

In C++, it's implementation-defined, and you truly get what you deserve.

Comments

• Anonymous
  April 21, 2004
  Did you mean "The second is that C# evaluates from left to right" ?
  
  
• Anonymous
  April 21, 2004
  Thanks, Wes
• Anonymous
  April 21, 2004
  I have some questions regarding the posers.
  
  In the second question, do the parens do anything to change the eval order? My guess is no. If I am incorrect, can you show me?
  
  Also, I am a little confused as to when the post increment operator gets executed. I always thought that it was after all processing in the current statement.
  
  Then in something like the following:
  
  z = 2;
  (1) z = z + z++;
  (2) z = z * z++;
  (3) z = z + (z++);
  (4) z = z * (z++);
  
  What would z contain for each (z=2 as the initial condition)?
  
  My answers would be 4, 4, 5, 6?
  Could you explain so that I can confirm my understanding?
  
  Thanks you guys.
• Anonymous
  April 21, 2004
  Jiho, z will result in 4 each time because C# seems to evaluate identifiers and their prefix/postfix operators from left to right regardless of how operator precedence determines they should be combined.
  
  I would expect 4, 4, 5, 6 as well from interpreting the C# spec. After having a long explaination from Wes in the previous blog I've been convinced that operator precedence doesn't necessarily control the order of identifer evaluation and the corresponding perfix/postfix operators.
  
  Eric, can you explain or give us a link to how C# relates precedence to actual evaluation?
  
  -Kael
• Anonymous
  April 22, 2004
  >>In C++, it's implementation-defined
  
  Correction: In C++, it is UNDEFINED.
  
  "implementation-defined" covers such things as "sizeof(int)" which are legal but different on different machines.
  
  "Undefined" means is illegal everywhere.
• Anonymous
  April 22, 2004
  Will someone define what exactly "implementation-defined" means?
• Anonymous
  April 22, 2004
  I could tell you what "implementation defined" means on this blog, but it might be different on Raymond Chen's.
  
• Anonymous
  April 23, 2004
  I agree with Kael. (3) and (4) seem at odds with 7.2.1 of the C# specification (specifically the last paragraph). I'd even go so far as to say it's a bug in the C# compiler.
  
  Any insight to describe why this behavior is (though I suspect the parens may be "optimized" out at compile time) would be appreciated.
• Anonymous
  April 23, 2004
  Parens are meaningful, but not for binary operators.
  
  I think the notion that taking:
  
  z++
  
  and writit it as:
  
  (z++)
  
  would convert it to
  
  ++z
  
  would be a fairly surprising bit of behavior. If you want ++z, why not just write ++z?
  
  
  
• Anonymous
  April 23, 2004
  I would never expect that (z++) would evaluate to ++z. Just because the increment may happen first, doesn't mean that the result of the expression would change. If z = 2 then (z++) would return 2 just as z++ would; the only difference with the parens would be that the operation would happen before any further evaluations of z.
  
  An explanation in detail is in a feedback to your original poser post: http://blogs.msdn.com/ericgu/archive/2004/04/19/116265.aspx#116801
• Anonymous
  April 26, 2004
  The comment has been removed
• Anonymous
  December 27, 2004
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