Is it legal/well-defined C++ to call a non-static method that doesn't access members through a null pointer?

I came across the following code recently:

class Foo
{
public:
    void bar();
    // .. other stuff
};

void Foo::bar()
{
    if(!this) {
        // .. do some stuff without accessing any data members
        return;
    }

    // .. do normal actions using data members
}

The code compiles because in C++ methods are just functions that are implicitly passed a pointer for 'this' and 'this' can be checked to be NULL just like any other pointer. Obviously this code is confusing and bad practice even though it doesn't crash; it would be pretty confusing to step through the code in the debugger, see a NULL pointer is about to have a method called on it and then not see the expected crash. My question is: does it violate the C++ standard to call SomeFooPtr->bar() where SomeFooPtr == NULL?

It occurs to me that it may not because the user defined operator-> returns a pointer, which means that even if that pointer is NULL it definitely hasn't been dereferenced (dereferencing a NULL pointer I'm sure is regarded by the sta开发者_运维技巧ndard as illegal or undefined). On the other hand the semantics of raw pointers don't necessarily have to match the semantics of user defined pointers -- perhaps operator-> on them is considered a dereference even though the compiler won't generate one.

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This will probably work on most systems, but it is Undefined Behaviour. Quoth the Standard:

5.2.5.3

If E1 has the type “pointer to class X,” then the expression E1->E2 is converted to the equivalent form (*(E1)).E2 [...]

And:

5.2.5.1

A postfix expression followed by a dot . or an arrow ->, optionally followed by the keyword template (14.8.1), and then followed by an id-expression, is a postfix expression. The postfix expression before the dot or arrow is evaluated;⁵⁸⁾ [...]

⁵⁸⁾ This evaluation happens even if the result is unnecessary to determine the value of the entire postfix expression, for example if the id-expression denotes a static member.

Evaluation of *x where x is a null pointer results in Undefined Behaviour, so yours is clearly a case of UB, before the function is even entered.

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This test is broken even if dereferencing was not an UB. It breaks when this-adjustments for multiple inheritance come into the play:

#include <stdio.h>
class B
{
    int value;
    public:
    void foo()
    {
        if (!this)
            printf("this==0\n");
        else 
            printf("safe\n");
    }
};
class A { public: int anotherValue; };
class Z : public A,public B {};

int main()
{
    Z *z=0;
    z->foo();
}

prints "safe" here.

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This is (all together now) undefined behavior. However, for many compilers it will work, with the additional constraint that the method must be non-virtual.

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It is UB. A good way to make it crash is to use it as a base class of a derived class that uses multiple inheritance. YMMV.

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It doesn't matter if it is legal, it is confusing to the reader. In the implementation where this code works, the vtable is being accessed by type, certainly not by object.

Moreover, I expect that this code was put in to cover for a constructor failure, which would mask a variety of problems elsewhere. Constructor failure should be handled correctly, not with a nasty kludge like the example.