C / C++ / MFC

28-Jun-05 21:28

I'm doing a little exercise after reading some chapter, it's about linked list, here is the description.
Create a struct that holds an int and a pointer to another
instance of the same struct. Write a function that takes
the address of one of these structs and an int indicating
the length of the list you want created(in my case: initialize). This function will make a whole chain of these structs (a linked list), starting
from the argument (the head of the list), with each one
pointing to the next. Make the new structs using new,
and put the count (which object number this is) in the int.
In the last struct in the list, put a zero value in the pointer
to indicate that it抯 the end. Write a second function(print) that
takes the head of your list and moves through to the end,
printing out both the pointer value and the int value for
each one.
After think for a while, I'm came up with a solution, but not a completed one.
The codes goes like this:

struct link<br />
{<br />
int a;<br />
link *next;<br />
};<br />
<br />
void initialize(link * head, int nCount)<br />
{<br />
	head->a = 0;<br />
	head->next = new link;<br />
	link *tmp = head->next;<br />
	tmp->a = 1;<br />
	for (int i = 2; i < nCount; i++)<br />
	{<br />
		link *lnk = new link;<br />
		tmp->next = lnk;<br />
		lnk->a = i;<br />
		tmp = tmp->next;<br />
	}<br />
}<br />
<br />
void print(link *head)<br />
{<br />
while (head->next) <br />
{<br />
cout << head->a << endl;<br />
head = head->next;<br />
}<br />
}<br />
<br />
int main()<br />
{<br />
link *head = new link;<br />
initialize(head, 10);<br />
print(head);<br />
}

I wonder how should I implement the initialize(...)?

Yulianto.28-Jun-05 21:47

28-Jun-05 21:47

I'm a bit rusty, so I'm not so sure if this works. Smile | :)

void initialize(link * head, int nCount)
{
	head->a = 0;
	for (int i = 1; i <= nCount; i++)
	{
		link *tmp = new link;
		if(i!=nCount)
		{
			tmp->a=i;
		}
		else
		{
			tmp->a=0;
		}
		head->next=tmp;
		head=tmp;
	}	
}

<italic>Work hard, Work effectively.

LiYS28-Jun-05 21:57

28-Jun-05 21:57

Yulianto. wrote:

void initialize(link * head, int nCount)
{
head->a = 0;
for (int i = 0; i < nCount; i++)
{
link *tmp = new link;
tmp->i+1;
head->next=tmp;
head=tmp;
}
}

Thanks but it didn't works out, and each link holds a invalid "a", in addition "head" is not head after the execution, I guess.

After testing, I found my initialize(...) seems to work, I still can't figure out exactly how? D'Oh! | :doh:

Yulianto.28-Jun-05 22:05

28-Jun-05 22:05

Oops , sorry, D'Oh! | :doh:

I've updated my post above. Try again.

<italic>Work hard, Work effectively.

LiYS28-Jun-05 22:14

28-Jun-05 22:14

Just back from run your initialize(...), the result is right!thanks, gotta know exactly how!Thanks again!

Yulianto.28-Jun-05 22:19

28-Jun-05 22:19

YongSheng Li wrote:
gotta know exactly how

You might want to google for 'Linked list'.

<italic>Work hard, Work effectively.

LiYS30-Jun-05 16:59

30-Jun-05 16:59

Thanks!I'll have a try!

...always look on the <blink>bright
side of life...
(Whistle)

LiYS28-Jun-05 22:43

28-Jun-05 22:43

Hi Yulianto:

I'm confused!in main()

<br />
int main()<br />
{<br />
 	link *head = new link;<br />
	initialize(head/*changed the head address inside*/, 10); <br />
	print(head/*still the same with initial head address, why???:doh: */); <br />
}<br />

Yulianto.28-Jun-05 22:56

28-Jun-05 22:56

Well, it's hard to explain. When you enter initialize, it didn't change the address of head in main. You should understand about pointer. Maybe this[^] could help.

<italic>Work hard, Work effectively.

John R. Shaw29-Jun-05 8:19

John R. Shaw

29-Jun-05 8:19

The head variable is passed to the function by value, which means that it is a copy of what is being passed. Therefore, you can use the variable head any way you want in the called function without changing the original variable in main.

Example:

void myFunc(int i)
{
    i = 25; // set copy equal to 25
}
int main()
{
    int i = 0; // local variable
    myFunc(i);
    // local variable i is still equal to 0
   return(0);
}

INTP
"The more help VB provides VB programmers, the more miserable your life as a C++ programmer becomes."
Andrew W. Troelsen

David Crow30-Jun-05 3:47

David Crow

30-Jun-05 3:47

How about something like:

struct link
{
    // default constructor
    link(){}
    
    link( const int n)
    {
        a = n;
        next = NULL;
    }
    
    int a;
 
    // pointer to next link in list
    link *next;
};
 
// pointers to first and last items in list
struct link *first = NULL;
struct link *last  = NULL;
 
//====================================================================
 
void addhead( struct link *newlink )
{
    if (NULL == last)
        last = newlink;
 
    if (NULL != first)
        newlink->next = first;
 
    first = newlink;
}
 
//====================================================================
 
void addhead( const int num )
{
    struct link *newlink = new struct link(num);
    addhead(newlink);
}
 
//====================================================================
 
void addtail( struct link *newlink )
{
    if (NULL == first)
        first = newlink;
 
    if (last != NULL)
        last->next = newlink;
    
    last = newlink;
}
 
//====================================================================
 
void addtail( const int num )
{
    struct link *newlink = new struct link(num);
    addtail(newlink);
}
 
//====================================================================
 
void main( void )
{
    struct link *list;
 
    // you can also put this in an initialize() function
    for (int x = 0; x < 10; x++)
        addtail(new struct link(x));
    
    // add a few other link (using a number)
    addtail(15);
    addhead(2);
    addtail(288);
 
    // print out the list
    list = first;
    while (NULL != list)
    {
        cout << list->a << endl;
        list = list->next;
    }
}

"Ideas are a dime a dozen. People who put them into action are priceless." - Unknown

LiYS30-Jun-05 17:00