Do you believe it's from our C++ exam?

[joehust]

Caculate the value of i after each step
int x=2, y=x+30;

struct A{

static int x;

int y;

public:

operator int( ){ return x+y; }

A operator ++(int){ return A(x++, y++); }

A(int x=::x+2, int y=::y+3){ A::x=x; A::y=y; }

int &h(int &x);

};

int &A::h(int &x)

{

for(int y=1; y!=1 || x<201; x+=11, y++)

if(x>200) { x-=21; y-=2;}

return x-=10;

}

int A::x=23;

void main( ){

A a(54, 3), b(65), c;

int i, &z=i, A::*p=&A::y;

i=b.x; //.................................¢Ù

z=a.x; //.................................¢Ú

i=c.*p; //.................................¢Û

i=a++; //.................................¢Ü

i=::x+c.y; //.................................¢Ý

i=a+b; //.................................¢Þ

b.h(i)=7; //.................................¢ß

}

and another one, write down the ouput of this program

#include <iostream.h>

struct A{A( ){ cout<<'A';}};

struct B{B( ){ cout<<'B';}};

struct C: A{C( ){ cout<<'C';}};
struct D: virtual B, C{D( ){ cout<<'D';}};

struct E: A{

C c;

E( ): c( ){ cout<<'E';}

};

struct F: virtual B, C, D, E{

F( ){ cout<<'F';}

};

void main( ){

A a; cout<<'\n';

B b; cout<<'\n';

C c; cout<<'\n';

D d; cout<<'\n';

E e; cout<<'\n';

F f; cout<<'\n';

}

[Victor Bazarov]

jo*****@gmail.com wrote:

Caculate the value of i after each step
[..]

Aside from the use of global variables (which has its place),
and 'void main', what's your objection? Is it too easy? Too hard?
Did you find that you could do it, but most couldn't? What?

and another one, write down the ouput of this program

[..]

Again, 'void main', and the use of an obsolete header <iostream.h>.
So? Did you solve it? It's not that difficult, and I am *sure* it
addresses just the topics your course contained.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[anon]

jo*****@gmail.com wrote:

struct A{

static int x;

int y;

public:

operator int( ){ return x+y; }

A operator ++(int){ return A(x++, y++); }

A(int x=::x+2, int y=::y+3){ A::x=x; A::y=y; }

int &h(int &x);

};

LOL @ struct A. It is written as x and y are private.

Easy exam, wasn't it?

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2007-07-03 09:54, anon wrote:

jo*****@gmail.com wrote:

>struct A{

static int x;

int y;

public:

operator int( ){ return x+y; }

A operator ++(int){ return A(x++, y++); }

A(int x=::x+2, int y=::y+3){ A::x=x; A::y=y; }

int &h(int &x);

};

LOL @ struct A. It is written as x and y are private.

Easy exam, wasn't it?

Maybe, but you failed. Since it's a struct all members are public by
default, so x and y are public, and the 'public:' is unnecessary.

--
Erik Wikström

[anon]

Erik Wikström wrote:

On 2007-07-03 09:54, anon wrote:

>jo*****@gmail.com wrote:

>>struct A{

static int x;

int y;

public:

operator int( ){ return x+y; }

A operator ++(int){ return A(x++, y++); }

A(int x=::x+2, int y=::y+3){ A::x=x; A::y=y; }

int &h(int &x);

};

LOL @ struct A. It is written as x and y are private.

Easy exam, wasn't it?

Maybe, but you failed. Since it's a struct all members are public by
default, so x and y are public, and the 'public:' is unnecessary.

I might have expressed myself wrongly, but read again what I wrote. To
me it looks like whoever wrote it expected x and y to be private.