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pointer arguments

Posted on 2000-03-14
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Last Modified: 2010-04-02
I have declared the following variable:

char * previous;

 I have allocated space for one chat to it:

previous = malloc(1);

I have initialized it as:

previous="\0";

I want to pass it to a function that will change its size using malloc and change its contents by copying another string to it:
                   strcpy(previous,curr);

 Now when I've returned from the function previous should be containing the string.

How am I going to pass the pointer to this function in order to achieve this???

I hope you all understand what I mean..
Thanx in advance..
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Question by:anemos
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6 Comments
 
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Accepted Solution

by:
KangaRoo earned 30 total points
ID: 2615550
I just answered that in the C section. hold on.
0
 

Author Comment

by:anemos
ID: 2615558
ooops! sorry for moving!
I just though that this is a more active area...
0
 
LVL 7

Expert Comment

by:KangaRoo
ID: 2615561
>> previous = malloc(1);
Better use new:
prev = new char[1];

>> previous="\0";

You've changed the pointer, not the contents of the previously allocated memory.
Might have wanted to do:

*previous = '\0';

You need to pass the pointer in a way that allows the function to change it, so pass the pointer by reference:

void change(char*& p)
{
   char* str = "hello";
   delete [] p;
   p = new char[strlen(str) + 1];
   strcpy(p, str);
}




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LVL 7

Expert Comment

by:KangaRoo
ID: 2615562
Never mind, such a thing rarely happens.
0
 

Author Comment

by:anemos
ID: 2615568
give me a minute to try it out..
0
 

Author Comment

by:anemos
ID: 2615576
Perfect! Quick and precise!
May god bless you and never let your machine crash!
 Thanx a lot,

 Nick
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