send pointer to function

Question

3.00/5 (2 votes)

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Hi I am trying to send an pointer to a function as following, but there is a problem with my code, can anybody help please:

void fun(int* p1, int* p2, int&s)
{
int size;
cin>>size;
s=size;
p1 = new int[size];
p2 = new int[size];
for(int i =0 ; i<size; ++i)
{
p1[i] = i;
p2[i] = i * i;
}
}
int main()
{
int* p1;
int* p2;
fun(p1,p2,size);
for(int i =0; i< size;++i)
cout<<p1[i]<<" "<<p2[i];
}

Posted 10-Jul-11 5:29am

sami984

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Comments

Richard MacCutchan 11-Jul-11 5:11am

I'm not sure why this question was down voted so I have 5'ed it to compensate. This is a perfectly reasonable question from someone who is trying to learn features of C++.

6 solutions

Solution 6

consider using this declaration:

oid fun(int* &p1, int* &p2, int&s)

because you need to derefernce the pointers

Posted 11-Jul-11 4:28am

KarstenK

Comments

Richard MacCutchan 11-Jul-11 11:12am

Another good suggestion, have a 5.

Solution 4

you didn't declare variable size in main() and variable s in fun()

C#

void fun(int* p1, int* p2, int&s)
{
   int size;
   cin>>s;     // replace size by s

   //s=size;         //comment or remove this line

   p1 = new int[size];
   p2 = new int[size];
   for(int i =0 ; i<size; ++i)
   {
      p1[i] = i;
      p2[i] = i * i;
   }
}

int main()
{

   int size = 0;             // Add this line 

   int* p1;
   int* p2;
   fun(p1,p2,size);
   for(int i =0; i< size;++i)
   cout<<p1[i]<<" "<<p2[i];
}

Copy and compile.

Posted 10-Jul-11 22:07pm

Ashish Tyagi 40

Updated 13-Jul-11 23:49pm

v6

Comments

CPallini 11-Jul-11 4:28am

size looks uninitialised to me.

Ashish Tyagi 40 11-Jul-11 4:32am

I am really sorry, Sir, let me modify it. And thinks for pointing me out..

Richard MacCutchan 11-Jul-11 4:36am

See also my answer, particularly my comments about p1 and p2. I guess we fixed half each! :)

Ashish Tyagi 40 11-Jul-11 4:48am

Ha ha ha you are right sir. +5 from me.. Ha ha ha :-)

Solution 1

You cant change the pointers p1, p2 to point to Heap as so, however, in the main function you have to intialize the size variable (need declare before using) before send it to the fun function. I think this help for you.

Posted 10-Jul-11 5:38am

Andrewpeter

Updated 10-Jul-11 5:39am

v2

Solution 2

size variable's scope is in function func only. In fact you are suppose to get a compilation error.
you need to declare size in main function too

Posted 10-Jul-11 5:42am

Mohibur Rashid

Solution 5

In addition to the above I'd like to add that the name you chose for your function could be misleading - it may not be funny at all! ;p

Posted 10-Jul-11 23:18pm

Stefan_Lang

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Richard MacCutchan · Accepted Answer · 2011-07-10T06:06:00

Solution 3

You should also realise that on return from fun you will be back to your original p1 and p2 which are uninitialised. To achieve what you are trying to do you should send p1 and p2 as references, or add another level of indirection thus:

void fun(int*& p1, int*& p2, int&s)
{
int size;
cin>>size;
s=size; // oops - restored this line
p1 = new int[size];
p2 = new int[size];

... // or

void fun(int** p1, int** p2, int&s)
{
int size;
cin>>size
s=size;
*p1 = new int[size];
*p2 = new int[size];
...

Posted 10-Jul-11 6:06am

Richard MacCutchan

Updated 10-Jul-11 21:58pm

v2

Comments

Espen Harlinn 10-Jul-11 14:57pm

Good points, my 5

Sergey Alexandrovich Kryukov 10-Jul-11 17:35pm

Yes, a 5.
--SA

Richard MacCutchan 11-Jul-11 4:05am

Thanks (and to Espen), but you missed my mistake which I have now corrected.

CPallini 11-Jul-11 4:26am

Simply the best.
The second form would be possibly: void fun(int ** pp1, int ** pp2, int * ps)

Richard MacCutchan 11-Jul-11 4:35am

Correct, I missed that part.

sami984 11-Jul-11 13:18pm

Thank u very much, its work now :)