The Meaning of Parentheses after Pointer

Question

I want to ask what does the (0) mean after the pointer i.e. Node* ptr1(0).

struct Node { string info; Node * next }; int main() { Node* ptr1 (0), *ptr2 (0), ptr1 = new Node; ptr2 = new Node; }

It should be noted that instead of initialising pointers with 0 or NULL one should use nullptr instead as of c++ 11. — sjrowlinson
– sjrowlinson, Commented May 20, 2016 at 15:05

Community · Accepted Answer · 2017-05-23 12:24:04Z

3

It means initializing the pointer with 0 or null.

Refer to this question for understanding the explanation behind it (they're actually the same; 0 and null in this context)

edited May 23, 2017 at 12:24

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answered May 20, 2016 at 15:00

Vucko

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3 Comments

Lightness Races in Orbit Over a year ago

0 and NULL are not the same.

rain_ Over a year ago

Also mind that that post has 7 years. You can also use nullptr

Vucko Over a year ago

Well of course they're not, but in this context @LightnessRacesInOrbit they represent the same...

Lightness Races in Orbit · Accepted Answer · 2016-05-20 15:28:35Z

For any integral type T, the following two declarations are effectively equivalent:

T obj(0); T obj = 0;

And since 0 is a null pointer constant, all you're doing here is initialising your two pointers to be null.

There are lots of ways to initialise objects, but consider how you declare objects of class type:

MyClass obj(someArguments...);

It's the same thing.

songyuanyao · Accepted Answer · 2016-05-20 15:31:07Z

This is direct initialization.

if T is a non-class type, standard conversions are used, if necessary, to convert the value of other to the cv-unqualified version of T.

For pointer type, initialize it with 0 makes it a null pointer. See Pointer conversions.

A null pointer constant (see NULL), can be converted to any pointer type, and the result is the null pointer value of that type.

Destructor · Accepted Answer · 2016-05-23 18:37:41Z

1

This simply initializes the pointer. And it is initializing it to be null.

answered May 23, 2016 at 18:37

Destructor

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Comments

Gregg · Accepted Answer · 2016-05-20 15:23:25Z

0

This is a constructor call. Since there is no constructor defined the compiler supplies one. After C++11 the preferred form is 'Node * ptr1 {0};' using braces.

answered May 20, 2016 at 15:23

Gregg

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2 Comments

Lightness Races in Orbit Over a year ago

There is no constructor for pointers, compiler-supplied or otherwise.

Gregg Over a year ago

Thanks, 'Node * ptr1 {nullptr};' would be a better also.

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