为什么这不对转换构造函数进行隐式转换？

Question

所以我有这个代码：

struct Foo {
    Foo() { cout << "default\n"; }
    Foo(const long long) { cout << "implicit\n"; }
};

struct Bar {
    Bar(const short param) : param(param) {}
    operator long long() const { return static_cast<long long>(param); }
    const short param;
};

我原以为 Foo foo = Bar(13) 会使用我的隐式转换，然后使用转换构造函数。 But it errors:

error: conversion from Bar to non-scalar type Foo requested

虽然这工作正常：Foo foo(Bar(13))。为什么我的隐式转换用于显式转换构造，而不用于隐式转换构造？

我从 https://en.cppreference.com/w/cpp/language/copy_initialization 得到的规则说：

The result of the conversion, which is a prvalue expression if a converting constructor was used, is then used to direct-initialize the object

Answer 1

首先从Bar到long long的隐式转换，从long long到Foo的隐式转换都是用户自定义转换。

Foo foo = Bar(13); 执行 copy initialization, the compiler will try to convert Bar to Foo implicitly. Two implicit conversions are required, i.e. converting Bar to long long and then converting long long to Foo. But only one user-defined conversion is allowed in one implicit conversion 序列。

Implicit conversion sequence consists of the following, in this order:

1) zero or one standard conversion sequence;

2) zero or one user-defined conversion;

3) zero or one standard conversion sequence.

A user-defined conversion consists of zero or one non-explicit single-argument constructor or non-explicit conversion function call

Foo foo(Bar(13)); 执行 direct initialization。将检查 Foo 的构造函数，并通过重载决策选择最佳匹配。只需要一个隐式的用户定义转换（从 Bar 到 long long）；之后调用Foo::Foo(long long)直接构造foo。

Answer 2

当你使用复制初始化然后根据documentation

In addition, the implicit conversion in copy-initialization must produce T directly from the initializer, while, e.g. direct-initialization expects an implicit conversion from the initializer to an argument of T's constructor.

重点是我的。事实并非如此。