Code Quiz!

[Bcm]

Hey guys, lets play a programming game!

Rules.
1. A programming problem is posted (I'll start)
2. 1st person to solve it with a detailed explanation, posts the next problem
3. The person who posted the problem confirms that solution is correct and passes the baton.
4. We all learn something.

 

[Bcm]

C++ Question. Stack Overflow when run! Why?

class Base
{
public:
   Base() {}
   virtual void func() { /* do something */ }
};

class Derived : public Base
{
public:
   Derived() {}
   virtual void func()
   {
     Base:func();
     /* do something else */
   }
};

void main()
{
   Derived d;
   d.func(); // Never returns!
}

 

[semaphore]

C++ Question. Stack Overflow when run! Why?

class Base
{
public:
   Base() {}
   virtual void func() { /* do something */ }
};

class Derived : public Base
{
public:
   Derived() {}
   virtual void func()
   {
     [b]Base::func();[/b]
     /* do something else */
   }
};

void main()
{
   Derived d;
   d.func(); // Never returns!
}

That will prevent function recursion. Cant remember why it does it thou.. someting todo with class labels..

 

[Bcm]

That will prevent function recursion. Cant remember why it does it thou..

Dude, I thought you were the C++ fanboy Jedi.

Ok, I'll explain...

Base:func(); is actually 2 lines
Base: -> Label called base
func(); -> Method call function

so Instead of using the correct line base::func(); which would call the correct method. The func() method is calling itself recursively until the stack overflows.

Your turn to ask a question.

 

[koeks]

C++ Question. Stack Overflow when run! Why?

class Base
{
public:
   Base() {}
   virtual void func() { /* do something */ }
};

class Derived : public Base
{
public:
   Derived() {}
   virtual void func()
   {
     Base:func();
     /* do something else */
   }
};

void main()
{
   Derived d;
   d.func(); // Never returns!
}

becuase you posted it... :

 

[koeks]

Dude, I thought you were the C++ fanboy Jedi.

Ok, I'll explain...

Base:func(); is actually 2 lines
Base: -> Label called base
func(); -> Method call function

so Instead of using the correct line base::func(); which would call the correct method. The func() method is calling itself recursively until the stack overflows.

Your turn to ask a question.

are programmers/developers this bored...

 

[semaphore]

Dude, I thought you were the C++ fanboy Jedi.

Ok, I'll explain...

Base:func(); is actually 2 lines
Base: -> Label called base
func(); -> Method call function

so Instead of using the correct line base::func(); which would call the correct method. The func() method is calling itself recursively until the stack overflows.

Your turn to ask a question.

Hey i got half of it right, it was cryptic.. ok let me think of one..

 

[semaphore]

argg ok i cant think of any.. ive only slept 3 hours today, you post more and let me try..

 

[Bcm]

are programmers/developers this bored...

Not really, today is hectic acctually. Nothing wrong with challenging yourself is there?

 

[semaphore]

void main(int argc, char* argv[])
{
  int n = 255;
  size_t size = sizeof(++n);
  printf("%d", n);
}

what will the output of that be without running the code.. ok that wasnt obscure..

 

[semaphore]

int main void()
{
   double x = 1.0;
   double n = 1e308;
   double d  = x/n;
   cout << "d = " d << endl;
   double e = pow(1.0+d,n);
   cout << "e = " << e;
   return 0;"
}

what is the output and why?

 

[Necuno]

int main void()
{
   double x = 1.0;
   double n = 1e308;
   double d  = x/n;
   cout << "d = " d << endl;
   double e = pow(1.0+d,n);
   cout << "e = " << e;
   return 0;"
}

what is the output and why?

{
";0 uɹnʇǝɹ   
;ǝ << " = ǝ" << ʇnoɔ   
;(u'p+0.1)ʍod = ǝ ǝ1qnop   
;1puǝ << p " = p" << ʇnoɔ   
;u/x =  p ǝ1qnop   
;803ǝ1 = u ǝ1qnop   
;0.1 = x ǝ1qnop   
}
()pıoʌ uıɐɯ ʇuı

 

[FarligOpptreden]

Feels like COS exams all over again...

 

[semaphore]

{
";0 uɹnʇǝɹ   
;ǝ << " = ǝ" << ʇnoɔ   
;(u'p+0.1)ʍod = ǝ ǝ1qnop   
;1puǝ << p " = p" << ʇnoɔ   
;u/x =  p ǝ1qnop   
;803ǝ1 = u ǝ1qnop   
;0.1 = x ǝ1qnop   
}
()pıoʌ uıɐɯ ʇuı

funny but no..

 

[wishblade]

int main void()
{
   double x = 1.0;
   double n = 1e308;
   double d  = x/n;
   cout << "d = " d << endl;
   double e = pow(1.0+d,n);
   cout << "e = " << e;
   return 0;"
}

what is the output and why?

yeah, may be a noob question, but what type of function declaration is "int main void()" anyway?

And then perhaps you will get a complaint about the extra " after the return statement...

 

[semaphore]

yeah, may be a noob question, but what type of function declaration is "int main void()" anyway?

And then perhaps you will get a complaint about the extra " after the return statement...

apart from the obvious typos its a 3rd year comp sci question, well i thats when i saw it.

 

[S1ght]

yeah, may be a noob question, but what type of function declaration is "int main void()" anyway?

And then perhaps you will get a complaint about the extra " after the return statement...

And don't forget the
cout << "d = " d << endl;
should be
cout << "d = "<< d << endl;

 

[wishblade]

Ah, I was wondering - thought I was seeing things... but the clearing of my potential stupid questions and lack of many years of experience with a "noob" reference still has effect

I'm still curious about the whole "int main void()" reference... Maybe I missed something along the way, but where does void fit in? Perhaps someone could help me out of my misery

 

[wishblade]

And don't forget the
cout << "d = " d << endl;
should be
cout << "d = "<< d << endl;

LOL, yeah, saw that one too after I had already posted. Perhaps the question should be re-posted correctly

 

[dequadin]

To me this is a math question:
x = 1.0;
n = 1e308 = 10^308
d = x/n
= 1.0
----
10^308
= 10^-308.
= 10e-308

e = (1.0 + 10^-308)^(10^308)
= 1.0^(10^308) + (10^-308) ^(10^308)
= 1.0 + 10^(-3080^308)
= 1.0 + 10^(-948640)

Now 10^(-948640) << 1 so we can assume 0 for this application.

e = 1.0;

QED