try to stump me on C++

TheCPUWizard, did I get this right:
http://www.codeguru.com/forum/showpo...9&postcount=51

If not, what the *&%! is the answer. :D

How about

Code:

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v[i] = ++i;

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Quote:

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Originally Posted by Zaccheus

Hmm, in that case I will have an educated guess and go back to my C roots, where all global data was, by default, initialised to zero. I would therefore think that C++ memsets such global data to zero before any user defined initialisation code is run. Therefore I would guess that X will already be zero when it is returned by f().

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That is PART of the answer. The second part of the answer has to deal with how the computer detects if statics are initialized. I dont have the spec with my, but it is (as one would assume) a "one-shot" deal.

Therefore the first time the static is referenced (or at some prior point) the assignment (initialization, or construction) must happen once. However if the static is referred to during this process it does NOT happen again.

Many people give the wrong answer that it is a recursive call that will stack crash....

I replied there... http://www.codeguru.com/forum/showth...=1#post1739503

Sorry for leaving you hanging...

Hmm, confirmed. CPUWiz is correct. Looked into the standards today just to confirm/refresh my memory. [basic.start.init - 3.6.2/1] :(

The value would be 0 and is well defined. Here the variable with static storage duration is initialized twice. Static initialization happens either to be via zero initialization or constant initialization (if a constant is used). Anything else is dynamic initialization and static initialization happens before any dynamic initialization. There is no constant expression involved, so zero initialization for the static variable happens and it gets the value 0. The function call then forms the dynamic initialization part, wherein, X is dynamically initialized with the statically initialized value of X i.e. 0. And hence, X is 0.

Pasting the code below for reference:

Code:

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static int X = f();  //dynamic initialization via the function call.
            //static initialization/zero initialization already has happened before the function call
int f() { return X; }  //effectively returns 0.
int main()
{
  cout << X << end;
}

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So ...

Code:

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#include <iostream>

int f();
static int X = f(); 

int f() { return X + 1; }

int main()
{
  cout << X << end;
}

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and

Code:

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#include <iostream>

int f();
static int X = f(); 

int f() { return X++; }

int main()
{
  cout << X << end;
}

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would be even more fun!

Quote:

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Originally Posted by JohnW@Wessex

Code:

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v[i] = i++;

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is equivelent to

Code:

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v[i] = i;
i++;

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Code:

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f(v[i],i++);

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Undefined.

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Well, they both are undefined. Actually, if you read a variable twice in an expression where you also write it, the result is undefined.

Btw, I wonder what happened about the interview. :rolleyes:

Quote:

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Originally Posted by Ejaz

Well, they both are undefined.

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Your link doesn't work.

http://www.research.att.com/~bs/bs_f...aluation-order

Code:

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v[ i ] = i++;

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For some reason I had thought that the right hand side of the assignment is always evaluated before the left hand side of the assignment, in which case the result would have been well defined.

Quote:

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Originally Posted by Ejaz

Well, they both are undefined.

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Yes, when I thought about it some more it makes sense.

Interestingly Visual Studio implements

v[i] = i++;

and

v[i] = ++i;

exactly as my initial gut reaction assumed they did.
Deliberate choice by Microsoft? Who knows.

I did read a comment on another forum that in C# these are well defined.

Quote:

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Originally Posted by JohnW@Wessex

I did read a comment on another forum that in C# these are well defined.

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Does that mean that in C# the right hand side is always evaluated before the left hand side (or is it perhaps the other way around)?

Wanders off to the C# forum ...

Quote:

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Originally Posted by Zaccheus

Wanders off to the C# forum ...

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It wasn't on this site.

You sure? I think he posted this in the C# forum
http://www.codeguru.com/forum/showthread.php?t=457023

Quote:

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Originally Posted by JohnW@Wessex

It wasn't on this site.

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I didn't assume it was, I just went there to ask the question on this site. :)

It was in one of the comments here.
Not having ever coded in C# I have no knowledge on this aspect.