C ++-测试空指针的最有效方法是什么？

Question

What is most efficient between the two ways of testing pointer nullity : if(pointer==NULL) or if(!pointer) .

MyObject* p;
[...]
// Solution 1
if ( p )
{ // Do something
}
// Solution 2
if ( p!=NULL )
{ // Do something
}

Answer 1

It makes no difference what so ever. It's purely a style issue what you prefer.

By the way, you should use nullptr rather than NULL if you use C++11 or later.

Answer 2

I prefer if (ptr) because:

1. It is short and clear
2. It doesn't depend on NULL keyword. Which have to be nullptr on C++11 or later as Jesper Juhl mentioned .
3. From SoapBox's comment on stackoverflow :

They are compatible with C++ classes such as auto_ptr that are objects that act as pointers and which provide a conversion to bool to enable exactly this idiom. For these objects, an explicit comparison to NULL would have to invoke a conversion to pointer which may have other semantic side effects or be more expensive than the simple existence check that the bool conversion implies.

Answer 3

Those are same. It makes no change in your program whatever you use.

Answer 4

I'd rather prefer the if (p!=NULL) as it prevents form accidental errors like casting the p in other thing, like an int for instance, between the declaration and the use of it.

Here the compiler should warn as int differ form NULL (initially NULL=(void*)0 )

Furthermore i prefer to declare like this :

MyObject* p=NULL;

because you cannot assure that p wont have a value different of zéro. If you miss to instantiate it between declaration and use.

And it is less misleading than the if (p==NULL){} else {} which may occur in an accidental assignment (only one =)

Off course in the C++11 or later this could involve some changes in your code even if the call of NULL is still working but deprecated.