Difference between (long) x and long(x)?

[mosimu]

I have never fully understood what the difference is between these two forms
of casting. Can anyone please clarify? long is a primitive data type so no,
it's not calling a class constructor. This is not simply a .NET feature
either, also works in VS6.

[Doug Harrison [MVP]]

mosimu wrote:

I have never fully understood what the difference is between these two forms
of casting. Can anyone please clarify? long is a primitive data type so no,
it's not calling a class constructor. This is not simply a .NET feature
either, also works in VS6.

Besides syntax, there is no difference. There's no difference when ctors are
involved, either. Of course, you can't do things like (char*) x using the
functional notation, and you can't creating temporaries using T() or T(x,y)
with the cast notation. But for single argument ctors, the functional and
cast notations are equivalent.

--
Doug Harrison
Microsoft MVP - Visual C++

[Ramsey]

For primitive types like long it doesn't matter. But if you had a class
foo then:

(foo) x calls x's conversion function.
foo (x) calls foo's contructor.

Ramsey

[Doug Harrison [MVP]]

Ramsey wrote:

For primitive types like long it doesn't matter. But if you had a class
foo then:

(foo) x calls x's conversion function.
foo (x) calls foo's contructor.

Nope. It's spelled out as I described in the C++ Standard, 5.2.3. The two
expressions you gave mean the same thing. Either can call a foo ctor or an x
conversion function. Here's an example:

struct X
{
operator int();
};

struct Y
{
Y(const X&);
};

void f(X x)
{
int i1 = (int) x;
int i2 = int(x);

Y y1 = (Y) x;
Y y2 = Y(x);
}
--
Doug Harrison
Microsoft MVP - Visual C++

[mosimu]

Thank you.

"mosimu" wrote:

I have never fully understood what the difference is between these two forms
of casting. Can anyone please clarify? long is a primitive data type so no,
it's not calling a class constructor. This is not simply a .NET feature
either, also works in VS6.