C++ scope on heap
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Why I am getting output as 10 after deleting it's scope? I have created a class A which has a pointer type member variable and another class B which has another member variable x. Now I am allocating memory on heap for B with value 10 for x. To create instance of class A, I have used memory address of x. Now I am deleting object b and printing value of pointer ref of class A.
I am getting why I am getting output as 10 as we can see that memory holding that value is no longer persisting on heap?
#include<iostream>
using namespace std;
class A
{
public:
int *ref;
A(int *ref):ref(ref)
{
}
};
class B{
public:
int x;
B(int x):x(x){}
};
int main()
{
B *b=new B(10);
A a(&b->x);
delete b;
cout<<*a.ref<<endl;
}
c++ scope heap heap-memory
asked Jan 4 at 7:45
Pankaj KumarPankaj Kumar
61
add a comment |
Why I am getting output as 10 after deleting it's scope? I have created a class A which has a pointer type member variable and another class B which has another member variable x. Now I am allocating memory on heap for B with value 10 for x. To create instance of class A, I have used memory address of x. Now I am deleting object b and printing value of pointer ref of class A.
I am getting why I am getting output as 10 as we can see that memory holding that value is no longer persisting on heap?
#include<iostream>
using namespace std;
class A
{
public:
int *ref;
A(int *ref):ref(ref)
{
}
};
class B{
public:
int x;
B(int x):x(x){}
};
int main()
{
B *b=new B(10);
A a(&b->x);
delete b;
cout<<*a.ref<<endl;
}
c++ scope heap heap-memory
asked Jan 4 at 7:45
Pankaj KumarPankaj Kumar
61
Nearly all of the delete operations in PC to not erase resources, but just free the space they occupied for future use. So the data might still exists for a certain amount of time, but the outcome of accessing that data is undefined.
– t.niese
Jan 4 at 7:53
1
Off topic but you are using the term scope in the wrong way. Scope refers to the part of the source code where a name is visible. It does not refer to the lifetime of a variable, that's usually called extent. So data allocated on the heap has indefinite extent (because it lives until you delete it) but any pointer variables pointing at that data have the normal scope rules.
– john
Jan 4 at 8:15
add a comment |
Why I am getting output as 10 after deleting it's scope? I have created a class A which has a pointer type member variable and another class B which has another member variable x. Now I am allocating memory on heap for B with value 10 for x. To create instance of class A, I have used memory address of x. Now I am deleting object b and printing value of pointer ref of class A.
I am getting why I am getting output as 10 as we can see that memory holding that value is no longer persisting on heap?
#include<iostream>
using namespace std;
class A
{
public:
int *ref;
A(int *ref):ref(ref)
{
}
};
class B{
public:
int x;
B(int x):x(x){}
};
int main()
{
B *b=new B(10);
A a(&b->x);
delete b;
cout<<*a.ref<<endl;
}
c++ scope heap heap-memory
asked Jan 4 at 7:45
Pankaj KumarPankaj Kumar
61
Why I am getting output as 10 after deleting it's scope? I have created a class A which has a pointer type member variable and another class B which has another member variable x. Now I am allocating memory on heap for B with value 10 for x. To create instance of class A, I have used memory address of x. Now I am deleting object b and printing value of pointer ref of class A.
I am getting why I am getting output as 10 as we can see that memory holding that value is no longer persisting on heap?
#include<iostream>
using namespace std;
class A
{
public:
int *ref;
A(int *ref):ref(ref)
{
}
};
class B{
public:
int x;
B(int x):x(x){}
};
int main()
{
B *b=new B(10);
A a(&b->x);
delete b;
cout<<*a.ref<<endl;
}
c++ scope heap heap-memory
c++ scope heap heap-memory
asked Jan 4 at 7:45
Pankaj KumarPankaj Kumar
61
asked Jan 4 at 7:45
Pankaj KumarPankaj Kumar
61
asked Jan 4 at 7:45
Pankaj KumarPankaj Kumar
61
asked Jan 4 at 7:45
Pankaj KumarPankaj Kumar
61
asked Jan 4 at 7:45
Pankaj KumarPankaj Kumar
61
61
Nearly all of the delete operations in PC to not erase resources, but just free the space they occupied for future use. So the data might still exists for a certain amount of time, but the outcome of accessing that data is undefined.
– t.niese
Jan 4 at 7:53
1
Off topic but you are using the term scope in the wrong way. Scope refers to the part of the source code where a name is visible. It does not refer to the lifetime of a variable, that's usually called extent. So data allocated on the heap has indefinite extent (because it lives until you delete it) but any pointer variables pointing at that data have the normal scope rules.
– john
Jan 4 at 8:15
add a comment |
Nearly all of the delete operations in PC to not erase resources, but just free the space they occupied for future use. So the data might still exists for a certain amount of time, but the outcome of accessing that data is undefined.
– t.niese
Jan 4 at 7:53
1
Off topic but you are using the term scope in the wrong way. Scope refers to the part of the source code where a name is visible. It does not refer to the lifetime of a variable, that's usually called extent. So data allocated on the heap has indefinite extent (because it lives until you delete it) but any pointer variables pointing at that data have the normal scope rules.
– john
Jan 4 at 8:15
Nearly all of the delete operations in PC to not erase resources, but just free the space they occupied for future use. So the data might still exists for a certain amount of time, but the outcome of accessing that data is undefined.
– t.niese
Jan 4 at 7:53
Nearly all of the delete operations in PC to not erase resources, but just free the space they occupied for future use. So the data might still exists for a certain amount of time, but the outcome of accessing that data is undefined.
– t.niese
Jan 4 at 7:53
1
1
Off topic but you are using the term scope in the wrong way. Scope refers to the part of the source code where a name is visible. It does not refer to the lifetime of a variable, that's usually called extent. So data allocated on the heap has indefinite extent (because it lives until you delete it) but any pointer variables pointing at that data have the normal scope rules.
– john
Jan 4 at 8:15
Off topic but you are using the term scope in the wrong way. Scope refers to the part of the source code where a name is visible. It does not refer to the lifetime of a variable, that's usually called extent. So data allocated on the heap has indefinite extent (because it lives until you delete it) but any pointer variables pointing at that data have the normal scope rules.
– john
Jan 4 at 8:15
add a comment |
3 Answers
3
active
oldest
votes
It's undefined behavior. Case in point, when I run your exmple in MSVC, my output is -572662307.
There's no guarantee on what happens if you try to access invalid memory. What will likely happen is that it accesses the invalid memory anyway, and if that memory hasn't changed, you may get the old value. That's actually a bad thing, because it appears that the program is working as intended, when in reality it isn't.
I am getting why I am getting output as 10 as we can see that memory
holding that value is no longer persisting on heap?
The memory is not valid anymore, but the number 10 may still be in that memory, and a.ref is still pointing to that memory. So even if the memory isn't valid, the 10 may still persist.
answered Jan 4 at 7:48
BlazeBlaze
7,6141832
add a comment |
What you have is undefined behavior because the program is accessing memory that does not belong to it anymore.delete frees the memory so that it can be reused, but that does not mean that value in that memory location is set to zero (if that is what you were expecting).
answered Jan 4 at 7:50
P.WP.W
18.7k41859
add a comment |
It is because of Dangling pointer effect. Although we have deleted the memory data is not yet erased or not other data being stored over there and our pointer is still pointing to that memory.
for reference
https://www.geeksforgeeks.org/dangling-void-null-wild-pointers/
answered Jan 4 at 9:31
Amit SharmaAmit Sharma
73
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
It's undefined behavior. Case in point, when I run your exmple in MSVC, my output is -572662307.
There's no guarantee on what happens if you try to access invalid memory. What will likely happen is that it accesses the invalid memory anyway, and if that memory hasn't changed, you may get the old value. That's actually a bad thing, because it appears that the program is working as intended, when in reality it isn't.
I am getting why I am getting output as 10 as we can see that memory
holding that value is no longer persisting on heap?
The memory is not valid anymore, but the number 10 may still be in that memory, and a.ref is still pointing to that memory. So even if the memory isn't valid, the 10 may still persist.
answered Jan 4 at 7:48
BlazeBlaze
7,6141832
add a comment |
It's undefined behavior. Case in point, when I run your exmple in MSVC, my output is -572662307.
There's no guarantee on what happens if you try to access invalid memory. What will likely happen is that it accesses the invalid memory anyway, and if that memory hasn't changed, you may get the old value. That's actually a bad thing, because it appears that the program is working as intended, when in reality it isn't.
I am getting why I am getting output as 10 as we can see that memory
holding that value is no longer persisting on heap?
The memory is not valid anymore, but the number 10 may still be in that memory, and a.ref is still pointing to that memory. So even if the memory isn't valid, the 10 may still persist.
answered Jan 4 at 7:48
BlazeBlaze
7,6141832
add a comment |
It's undefined behavior. Case in point, when I run your exmple in MSVC, my output is -572662307.
There's no guarantee on what happens if you try to access invalid memory. What will likely happen is that it accesses the invalid memory anyway, and if that memory hasn't changed, you may get the old value. That's actually a bad thing, because it appears that the program is working as intended, when in reality it isn't.
I am getting why I am getting output as 10 as we can see that memory
holding that value is no longer persisting on heap?
The memory is not valid anymore, but the number 10 may still be in that memory, and a.ref is still pointing to that memory. So even if the memory isn't valid, the 10 may still persist.
answered Jan 4 at 7:48
BlazeBlaze
7,6141832
It's undefined behavior. Case in point, when I run your exmple in MSVC, my output is -572662307.
There's no guarantee on what happens if you try to access invalid memory. What will likely happen is that it accesses the invalid memory anyway, and if that memory hasn't changed, you may get the old value. That's actually a bad thing, because it appears that the program is working as intended, when in reality it isn't.
I am getting why I am getting output as 10 as we can see that memory
holding that value is no longer persisting on heap?
The memory is not valid anymore, but the number 10 may still be in that memory, and a.ref is still pointing to that memory. So even if the memory isn't valid, the 10 may still persist.
answered Jan 4 at 7:48
BlazeBlaze
7,6141832
edited Jan 4 at 8:39
answered Jan 4 at 7:48
BlazeBlaze
7,6141832
answered Jan 4 at 7:48
BlazeBlaze
7,6141832
answered Jan 4 at 7:48
BlazeBlaze
7,6141832
7,6141832
add a comment |
add a comment |
What you have is undefined behavior because the program is accessing memory that does not belong to it anymore.delete frees the memory so that it can be reused, but that does not mean that value in that memory location is set to zero (if that is what you were expecting).
answered Jan 4 at 7:50
P.WP.W
18.7k41859
add a comment |
What you have is undefined behavior because the program is accessing memory that does not belong to it anymore.delete frees the memory so that it can be reused, but that does not mean that value in that memory location is set to zero (if that is what you were expecting).
answered Jan 4 at 7:50
P.WP.W
18.7k41859
add a comment |
What you have is undefined behavior because the program is accessing memory that does not belong to it anymore.delete frees the memory so that it can be reused, but that does not mean that value in that memory location is set to zero (if that is what you were expecting).
answered Jan 4 at 7:50
P.WP.W
18.7k41859
What you have is undefined behavior because the program is accessing memory that does not belong to it anymore.delete frees the memory so that it can be reused, but that does not mean that value in that memory location is set to zero (if that is what you were expecting).
answered Jan 4 at 7:50
P.WP.W
18.7k41859
answered Jan 4 at 7:50
P.WP.W
18.7k41859
answered Jan 4 at 7:50
P.WP.W
18.7k41859
answered Jan 4 at 7:50
P.WP.W
18.7k41859
18.7k41859
add a comment |
add a comment |
It is because of Dangling pointer effect. Although we have deleted the memory data is not yet erased or not other data being stored over there and our pointer is still pointing to that memory.
for reference
https://www.geeksforgeeks.org/dangling-void-null-wild-pointers/
answered Jan 4 at 9:31
Amit SharmaAmit Sharma
73
add a comment |
It is because of Dangling pointer effect. Although we have deleted the memory data is not yet erased or not other data being stored over there and our pointer is still pointing to that memory.
for reference
https://www.geeksforgeeks.org/dangling-void-null-wild-pointers/
answered Jan 4 at 9:31
Amit SharmaAmit Sharma
73
add a comment |
It is because of Dangling pointer effect. Although we have deleted the memory data is not yet erased or not other data being stored over there and our pointer is still pointing to that memory.
for reference
https://www.geeksforgeeks.org/dangling-void-null-wild-pointers/
answered Jan 4 at 9:31
Amit SharmaAmit Sharma
73
It is because of Dangling pointer effect. Although we have deleted the memory data is not yet erased or not other data being stored over there and our pointer is still pointing to that memory.
for reference
https://www.geeksforgeeks.org/dangling-void-null-wild-pointers/
answered Jan 4 at 9:31
Amit SharmaAmit Sharma
73
answered Jan 4 at 9:31
Amit SharmaAmit Sharma
73
answered Jan 4 at 9:31
Amit SharmaAmit Sharma
73
answered Jan 4 at 9:31
Amit SharmaAmit Sharma
73
73
add a comment |
add a comment |
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Nearly all of the delete operations in PC to not erase resources, but just free the space they occupied for future use. So the data might still exists for a certain amount of time, but the outcome of accessing that data is undefined.
– t.niese
Jan 4 at 7:53
1
Off topic but you are using the term scope in the wrong way. Scope refers to the part of the source code where a name is visible. It does not refer to the lifetime of a variable, that's usually called extent. So data allocated on the heap has indefinite extent (because it lives until you delete it) but any pointer variables pointing at that data have the normal scope rules.
– john
Jan 4 at 8:15