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C++ Discussion Thread
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May 10th, 2009

even this simple code?? c'mon c++

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I have this code:
in my judgement it should print 1 and 2, instead it prints 0 and 1
some advice pls... Im using dev c++

C++ Syntax (Toggle Plain Text)
#include <iostream>
 
class A
{
public:
A()
{
arr = new int[2];
}
~A(){ delete [] arr; }
 
int * arr ;
};
 
 
int main() 
{
int * n ; 
 
{//induced code block
  A aObj;
aObj.arr[0] = 1;
aObj.arr[1] = 2;
n = aObj.arr;
 
/*
 BTW I assume n now points to the same add of  aObj.arr ... am I wrong??
*/
}
 
std::cout << "\n" << (*n) << " " << (*n+1);
system("pause");
}
#include <iostream>

class A
{
public:
A()
{
arr = new int[2];
}
~A(){ delete [] arr; }

int * arr ;
};


int main() 
{
int * n ; 

{//induced code block
  A aObj;
aObj.arr[0] = 1;
aObj.arr[1] = 2;
n = aObj.arr;

/*
 BTW I assume n now points to the same add of  aObj.arr ... am I wrong??
*/
}

std::cout << "\n" << (*n) << " " << (*n+1);
system("pause");
}

Last edited by Ancient Dragon; May 11th, 2009 at 12:20 am. Reason: add line numbers

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namehere05

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May 11th, 2009

Re: even this simple code?? c'mon c++

The problem is scoping -- object A goes out of scope before the cout statement on line 31. Delete the { and } on lines 20 and 29 and your program will work as you expect it to.

Ancient Dragon

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May 11th, 2009

Re: even this simple code?? c'mon c++

good, thanks I should've seen that
the morale is dont let garbage printed on the screen fool you even if it looks extremly similar too what you expect

Last edited by namehere05; May 11th, 2009 at 12:28 am.

namehere05

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May 11th, 2009

Re: even this simple code?? c'mon c++

Keep in mind that this line (see red part):

(Toggle Plain Text)

std::cout << "\n" << (*n) << " " << (*n+1);

displays aObj.arr[0] + 1 , not aObj.arr[1] . In your case, they are the same, but if you change line 23 to equal something other than 2, you'll still get 1 2 as output.

Last edited by VernonDozier; May 11th, 2009 at 12:33 am.

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