Why does this compile?

[Alexander Pandit]

here's the code :

using namespace std;

class A {
public:
A(int j): i(j) {
cout << "constructing an A object :" << i << endl;
}

~A() {
cout << "destructing an A object " << i << endl;
}

void show() {
cout << i << endl;
}

private:
int i;
};

int main() {

A *a;
a->show(); // <--- shouldn't this give a compile error , or does this
result in undefined behaviour?
return 0;
}

 

[Mike Wahler]

here's the code :

#include <iostream>

using namespace std;

class A {
public:
A(int j): i(j) {
cout << "constructing an A object :" << i << endl;
}

~A() {
cout << "destructing an A object " << i << endl;
}

void show() {

void show() const {

cout << i << endl;
}

private:
int i;
};

int main() {

A *a;
a->show(); // <--- shouldn't this give a compile error , or does this
result in undefined behaviour?

Undefined behavior. No diagnostic is required (but neither is
one prohibited).

return 0;
}

I try to prevent such errors by *always* initializing any objects
created.

-Mike

 

[Alexander Pandit]

Do you mean i should initialize it like :

A *a = NULL;
a->show(); // <-- Then I get a runtime-error

-Alex

 

[angelo]

here's the code :

using namespace std;

class A {
public:
A(int j): i(j) {
cout << "constructing an A object :" << i << endl;
}

~A() {
cout << "destructing an A object " << i << endl;
}

void show() {
cout << i << endl;
}

private:
int i;
};

int main() {

A *a;
a->show(); // <--- shouldn't this give a compile error , or does this
result in undefined behaviour?
return 0;
}

If you added a line:
a = new A(1);
the following statements would seem absolutely natural and correct,
right? But the point is the C/C++ doesn't check if a pointer is assigned
a valid value or not, so as far as syntax is concerned, it has no
error at all.
Moreover, as you have said, the result is undefined, so the compiler's
not generating an error message is just natural.

 

[Mike Wahler]

Do you mean i should initialize it like :

A *a = NULL;

No, initialize it with the address of a type 'A'
object. If the object doesn't exist yet, don't
define the pointer until it does. In any case,
in the context you showed, I see no reason
to use a pointer at all anyway. Just define
the object directly:

A a(42);
a.show();

a->show(); // <-- Then I get a runtime-error

You may or may not get a 'run-time error'.
The behavior is still undefined.

BTW please don't top-post. Thanks.

-Mike

 

[Alexander Pandit]

thank you.

-alex

No, initialize it with the address of a type 'A'
object. If the object doesn't exist yet, don't
define the pointer until it does. In any case,
in the context you showed, I see no reason
to use a pointer at all anyway. Just define
the object directly:

A a(42);
a.show();


You may or may not get a 'run-time error'.
The behavior is still undefined.

BTW please don't top-post. Thanks.

-Mike

 

[Alexander Pandit]

BTW please don't top-post. Thanks.

-Mike

Ups didn't know until now what top-/bottom posting means. Now I do and won't
top-post again.

-alex

 

[Ron Natalie]

Do you mean i should initialize it like :

A *a = NULL;
a->show(); // <-- Then I get a runtime-error

-Alex

Not necessarily. Even the above code is undefined behavior
(might get a runtime error, might just hose up some other
memory in your program).

A* a = NULL;

if(a) a->show(); //

or intialize a with the address of an actual object.

 

[Karthik Kumar]

here's the code :

using namespace std;

class A {
public:
A(int j): i(j) {
cout << "constructing an A object :" << i << endl;
}

~A() {
cout << "destructing an A object " << i << endl;
}

void show() {
cout << i << endl;
}

private:
int i;
};

int main() {

A *a;
a->show(); // <--- shouldn't this give a compile error , or does this
result in undefined behaviour?
return 0;
}

This is *undefined behaviour*. The compiler would not flag an error
since this is perfectly valid as per the C++ language syntax.
But it can definitely flag a warning since we are dealing with in an
uinitialized local variable here. Set your compiler warning level
at its highest level and it *might* catch this one.