Constructor Initialization of primitive data types in CPP

Question

In cpp, we can have primitive data-types initialized as

int a(32);

How does this constructor initialisation work? Does C++ treat it as an object?

What exact do you mean by "treats it as an object"? When it gets compiled it will move the literal constant 32 to the variables memory slot. Nothing more. So its just a syntax thing. — d_inevitable
– d_inevitable, Commented Jun 1, 2012 at 2:40
No, in C++ the primitives are not treated as objects. As far as integers are concerned, this is just an alternative syntax for a=32. — Sergey Kalinichenko
– Sergey Kalinichenko, Commented Jun 1, 2012 at 2:41

Alok Save · Accepted Answer · 2012-06-01 02:46:28Z

3

This is best described in:

C++03 8.5 Initializers
Para 12 & 13:

.......
The initialization that occurs in new expressions (5.3.4), static_cast expressions (5.2.9), functional notation type conversions (5.2.3), and base and member initializers (12.6.2) is called
direct-initialization and is equivalent to the form

T x(a);

If T is a scalar type, then a declaration of the form
T x = { a };
is equivalent to
T x = a;

In the question the type is int which is a scalar type.

answered Jun 1, 2012 at 2:46

Alok Save

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Reinier Torenbeek · Accepted Answer · 2012-06-01 02:42:39Z

2

This is so-called direct-initialization. In C++, integers are not objects and the expression that you write here is not a constructor. It just initializes a to the value of 32.

answered Jun 1, 2012 at 2:42

Reinier Torenbeek

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3 Comments

vagrawal13 Over a year ago

Got confused by the statement "The other way to initialize variables, known as constructor initialization, is done by enclosing the initial value between parentheses (()):" given here at cplusplus.com/doc/tutorial/variables

Jerry Coffin Over a year ago

@vagrawal13: Yet another reason to avoid cplusplus.com.

Reinier Torenbeek Over a year ago

@vagrawal13: that statement is somewhat confusing indeed. For scalar types, the difference between the two different kinds of initialization is syntactic only. For objects of class type, the situation is more subtle. See Is there a difference in C++ between copy initialization and assignment initialization? for a good read.

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