Linked List and Pointers in the if statements
Hi I am studying basic Linked List right now. I am reading a related book and got stuck in this line.
"head" is pointer that was declared as a private member variable of another class called "NumberList" and will be initialized to nullptr by its constructor.
My question is what does pointer variable with "!" operator mean ?
In the book, it says "Because 'head' is a null pointer, the condition '!head' is true." How does this if statement work?
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"head" is pointer that was declared as a private member variable of another class called "NumberList" and will be initialized to nullptr by its constructor.
My question is what does pointer variable with "!" operator mean ?
In the book, it says "Because 'head' is a null pointer, the condition '!head' is true." How does this if statement work?
Last edited on
nullptr is basically an alias for 0, except it's "pointer-like". So
is equivalent to
In fact, 0 was the preferred value to use for null pointers before nullptr came around.
Now, checking equality to zero is the same as comparing to false, so
is equivalent to
Finally, logically x == false is the same predicate as !x. (!x) == (x == false), no matter the value of x. So
is equivalent to
if (head == nullptr)is equivalent to
if (head == 0)In fact, 0 was the preferred value to use for null pointers before nullptr came around.
Now, checking equality to zero is the same as comparing to false, so
if (head == 0)is equivalent to
if (head == false)Finally, logically x == false is the same predicate as !x. (!x) == (x == false), no matter the value of x. So
if (head == false)is equivalent to
if (!head)
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