What are different ways of initialization

Question

What is the difference between:

long long int flag=n-1; and long long int flag(n-1);

Are these things same?I have seen the latter couple of time but have no proper idea about it.

They're the same for scalar types. In general they're different. — Brian Bi, Mar 04 '15 at 18:26
"In general they're different"..what do you mean by this statement?what are the situations they are different at? — chota bheem, Mar 04 '15 at 18:29
@LogicStuff Actually i didn't know what it is called thats why i asked here.I searched different ways of initialization but that doesn't helped me — chota bheem, Mar 04 '15 at 18:32

score 6 · Accepted Answer · answered Mar 04 '15 at 18:28

6

long long int flag = n-1 is copy initialization. For class types, this only considers non-explicit constructors and user-defined conversions.

long long int flag(n-1) is direct initialization. This considers all constructors and user-defined conversions.

However, these differences only matter for class types. For fundamental types, there is no difference.

answered Mar 04 '15 at 18:28

Pradhan

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That was pretty clear Mister..Thanks – chota bheem Mar 04 '15 at 18:38

score 4 · Answer 2 · answered Mar 04 '15 at 18:27

Strictly speaking, the first one is called "copy initialization" and the second one is called "direct initialization". For primitive types, there is no difference in code behavior.

More details can be found at http://en.cppreference.com/w/cpp/language/initialization.

score 0 · Answer 3 · answered Mar 04 '15 at 18:29

For built-in types like the ones you listed, there is no difference.

But if you were dealing with objects, then the first would call the assignment operator, and the second would call the copy constructor.

So if you had overloaded the assignment operator and copy constructor, then

operator=(const Object& rhs)

would be called by the first, and

Object(const Object& rhs)

would be called by the second.

It's important to note this because of the differences in the way these functions behave. The copy constructor of a derived object will always call all the base objects' constructors. But creating an object will always automatically call its constructor anyway, and this will automatically call the base constructors also.

So, for instance, if you created an object and then assigned it

Object o;
o = n - 1;

Then you would be calling the object's constructor, all of its base constructors, and then its assignment operator. This is why it's generally best to use the copy constructor for objects. For built-in types it doesn't matter.

What are different ways of initialization

3 Answers3