Accessing the memory locations created on heap

Question

Why the following C++ code does not give a segmentation fault, I am trying to access something that I have deleted.

#include <iostream>
using namespace std;
void fun ( int * ptr )
{
    delete ptr;
}
int main ()
{
    int * ptr = new int ;
    *ptr = 6;
    fun ( ptr );
    cout<<*ptr;
    return 0;
 }

It is one of my assignment questions that I got on Segmentation fault — Nilesh Kumar, Mar 22 '15 at 07:37
Please check : http://stackoverflow.com/questions/2346806/what-is-segmentation-fault — Tharif, Mar 22 '15 at 07:47

score 1 · Accepted Answer · answered Mar 22 '15 at 07:41

1

Accessing something you've deleted doesn't automatically result in segfault.

The behavior is undefined. It might segfault, it might not. You can never know.

answered Mar 22 '15 at 07:41

Emil Laine

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How can you guarantee that it will not cause segfault ? – Tharif Mar 22 '15 at 07:47
@utility Because the C++ standard defines it to be *undefined behavior*. It doesn't say it should cause a segfault. Segfault only happens if the OS is not happy with you trying to access the memory address. – Emil Laine Mar 22 '15 at 07:57

Accessing the memory locations created on heap

1 Answers1