c++ - 像 "Foo(12,3);"这样的未命名变量的强制构造仍然是声明符吗？

Question

考虑以下简单示例:

struct A 
{
    int a;
    A(int b);
};

A *p;

A::A(int b) : a(b)
{
    p = this;
    std::cout << "A()" << std::endl; 
} 



int main()
{
    A(10); //What is it?
    std::cout << p -> a << std::endl; //10
}

我不确定 N3797::8.5 [dcl.init] 是否可以应用在这里，因为 N3797::8.5/1 [dcl.init] 说:

A declarator can specify an initial value for the identifier being declared. The identifier designates a variable being initialized

在没有声明者的情况下。这意味着该规则不适用。

如果它只是一个构造函数调用，它在哪里指定 int 标准，即 class-name(argument_list) 形式的表达式分配合适数量的内存(哪个分配函数接管了它？)和初始化一个对象？我希望获得有关此类表达式如何工作的更多详细信息？

Answer 1

A(10);

这是一个丢弃值表达式。根据 [expr.type.conv]，

A simple-type-specifier (7.1.6.2) or typename-specifier (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).

'对应的转换表达式'是(A)10 ，这又相当于 static_cast<A>(10) ([expr.cast]/4)，以及 static_cast定义为 [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.

这就是你的声明符。