Initialising smart pointer

Question

Possible Duplicate:
Do the parentheses after the type name make a difference with new?

Whats the difference between the following initialisations? In the tutorial, it is as in case #1 but does it make any difference if i use the #2 way below?

struct X
{
    X() {}
    int x;
};

int main()
{
    std::auto_ptr<X> p1(new X);   // #1
    std::auto_ptr<X> p2(new X()); // #2
}

See this good reference: http://stackoverflow.com/questions/620137/do-the-parentheses-after-the-type-name-make-a-difference-with-new — JRL, Feb 05 '12 at 02:36

score 1 · Accepted Answer · edited May 23 '17 at 10:34

The smart pointer doesn't make any difference here. Both smart pointers are initialized in the same way, with a pointer to X. The difference is how X is initialized. If there is a difference and what the difference is depends on how X is defined. This answer has an excellent description of what happens in different cases. In this case since X has a default constructor they get initialized the same. However if there were no default constructor they would be initialized differently.

Initialising smart pointer

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