Two questions about lifetime and scope.

[Jason Heyes]

Consider the function with this prototype:

void foo(Bar bar);

1. Can a program call foo and allow only one addressable Bar object to exist
at a time?
Now consider the following program:

class Bar
{
int value;
public:
Bar() : value(0) { }
void change(int new_value) { value = new_value; }
};

void foo(Bar bar) { }

int main()
{
Bar bar;
bar.change(1);
foo(bar);
return 0;
}

2. Can this program be modified without change to foo and Bar so that:

i) main calls foo with the same value of bar, and
ii) only one addressable Bar object exists at a time.
Please support your answers with examples. Any help is appreciated.

[Gernot Frisch]

"Jason Heyes" <ja********@optusnet.com.au> schrieb im Newsbeitrag
news:41***********************@news.optusnet.com.a u...

Consider the function with this prototype:

void foo(Bar bar);

1. Can a program call foo and allow only one addressable Bar object
to exist
at a time?
Hm... I don't see any way now.
2. Can this program be modified without change to foo and Bar so
that:

i) main calls foo with the same value of bar, and

class Bar
{
public:
void foo() {this->foo_something();}
}

Now, you need a Bar to 'foo'.

ii) only one addressable Bar object exists at a time.

int IsOneBarThere=0;
class Bar
{
Bar() {if (IsOneBarThere++) assert(0);}
}

What exaclty are you trying to do?

[Alf P. Steinbach]

* Jason Heyes:

[homework questions]

HOMEWORK is not answered here.

Previously you posted the same two questions with a third
question,

3. What is the practical significance, if any, of the previous two
questions in relation to data sharing and copy-on-write?

which says this is HOMEWORK.

Go away.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

[Gernot Frisch]

"Alf P. Steinbach" <al***@start.no> schrieb im Newsbeitrag
news:41*****************@news.individual.net...

* Jason Heyes:

[homework questions]

HOMEWORK is not answered here.

Previously you posted the same two questions with a third
question,

3. What is the practical significance, if any, of the previous two
questions in relation to data sharing and copy-on-write?

which says this is HOMEWORK.

Oh. I've not seen this, but my solution uses a global variable. I'd
like to see the face of your professor when you show this... :)
-Gernot

[Jason Heyes]

"Alf P. Steinbach" <al***@start.no> wrote in message
news:41*****************@news.individual.net...

* Jason Heyes:

[homework questions]

HOMEWORK is not answered here.

Previously you posted the same two questions with a third
question,

3. What is the practical significance, if any, of the previous two
questions in relation to data sharing and copy-on-write?

which says this is HOMEWORK.

Go away.

Stop saying all my posts are homework. You've made everyone paranoid.

[Rolf Magnus]

Gernot Frisch wrote:

"Jason Heyes" <ja********@optusnet.com.au> schrieb im Newsbeitrag
news:41***********************@news.optusnet.com.a u...

Consider the function with this prototype:

void foo(Bar bar);

1. Can a program call foo and allow only one addressable Bar object
to exist
at a time?

Hm... I don't see any way now.

2. Can this program be modified without change to foo and Bar so
that:

i) main calls foo with the same value of bar, and

class Bar
{
public:
void foo() {this->foo_something();}
}

Now, you need a Bar to 'foo'.

ii) only one addressable Bar object exists at a time.

int IsOneBarThere=0;
class Bar
{
Bar() {if (IsOneBarThere++) assert(0);}
}

What exaclty are you trying to do?

That code does not follow the requirement of not changing foo and Bar.

[Jason Heyes]

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:cn************@news.t-online.com...

Gernot Frisch wrote:

"Jason Heyes" <ja********@optusnet.com.au> schrieb im Newsbeitrag
news:41***********************@news.optusnet.com.a u...

Consider the function with this prototype:

void foo(Bar bar);

1. Can a program call foo and allow only one addressable Bar object
to exist
at a time?

Hm... I don't see any way now.

2. Can this program be modified without change to foo and Bar so
that:

i) main calls foo with the same value of bar, and

class Bar
{
public:
void foo() {this->foo_something();}
}

Now, you need a Bar to 'foo'.

ii) only one addressable Bar object exists at a time.

int IsOneBarThere=0;
class Bar
{
Bar() {if (IsOneBarThere++) assert(0);}
}

What exaclty are you trying to do?

That code does not follow the requirement of not changing foo and Bar.

This is correct.