c++ - 函数样式转换与调用构造函数

Question

如果我有一个类 A，并且我写了 A(5);，它显然是一个临时变量。

但是不清楚 A(5); 是构造函数调用(使用 5 作为参数)，还是函数样式转换，转换 5 到 A。有人可以给我解释一下吗？

Answer 1

这是一种函数式类型转换，它创建了一个 t来自 int通过调用构造函数。在 C++ 中无法显式调用构造函数。

这在 [expr.type.conv] 中有描述(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.

自 t是一个简单类型说明符，这等同于相应的强制转换表达式。这允许执行相当于 static_cast 的操作。 ( [expr.cast]/4 )，它定义了转换的最终结果:

[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.