Function style casting vs calling constructor

Question

If I have a class A, and I write A(5);, it clearly makes a temporary variable.

But what is not clear if A(5); is a constructor call (using 5 as parameter), or if this is a function style cast, casting 5 to A. Can someone explain it to me please?

I do not understand, what is not clear for you. Casting an `int` to object `t` would look like `(t) 5;`. — Zereges, May 16 '16 at 13:37
As the famous russian saying, goes, 'smack in the forehead is the same as the smack to the forehead'. — SergeyA, May 16 '16 at 13:40
@RSahu http://stackoverflow.com/questions/4474933/what-exactly-is-or-was-the-purpose-of-c-function-style-casts — Rakete1111, May 16 '16 at 13:40
Maybe [this](http://stackoverflow.com/questions/3484371/whats-the-difference-between-functionmyvar-and-functionmyvar) will make it more clear. — rozina, May 16 '16 at 13:59
I don't think the marked question is a duplicate: this is about function-style conversions vs. explicit constructor calls. Cast-notation isn't in the question. — TartanLlama, May 16 '16 at 14:47

TartanLlama · Accepted Answer · 2016-05-16T13:57:48.590

It's a functional-style type conversion which creates a t from an int by calling the constructor. There is no way to explicitly call a constructor in C++.

This is described in [expr.type.conv] (N3337):

5.2.3 Explicit type conversion (functional notation)

1) A simple-type-specifer (7.1.6.2) or typename-specifer (14.6) followed by a parenthesized expression-list constructs a value of the specified type given the expression list. If the expression list is a single expression, the type conversion expression is equivalent (in definedness, and if defined in meaning) to the corresponding cast expression (5.4). If the type specified is a class type, the class type shall be complete. If the expression list specifies more than a single value, the type shall be a class with a suitably declared constructor (8.5, 12.1), and the expression T(x1, x2, ...) is equivalent in effect to the declaration T t(x1, x2, ...); for some invented temporary variable t, with the result being the value of t as a prvalue.

Since t is a simple-type-specifier, this is equivalent to the corresponding cast expression. This is allowed to carry out the equivalent of a static_cast ([expr.cast]/4), which defines the final result of the conversion:

[expr.static.cast]/4: Otherwise, an expression e can be explicitly converted to a type T using a static_cast of the form static_cast<T>(e) if the declaration T t(e); is well-formed, for some invented temporary variable t (8.5). The effect of such an explicit conversion is the same as performing the declaration and initialization and then using the temporary variable as the result of the conversion. The expression e is used as a glvalue if and only if the initialization uses it as a glvalue.

Please yes if it is possible – andre3312 May 16 '16 at 13:44 — andre3312, May 16 '16 at 13:44

Function style casting vs calling constructor

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