c++ - boost 变体 : Is there magic in the binding order?

Question

我一直在阅读我能找到的关于 Boost Variant 的所有内容. (我很头疼。)有些人对分配字符串文字被保存为 bool 值感到惊讶。如果我在 bool 之前(之后？)列出了 char*，字符串文字会被保存为 char* 字符串吗？ v1 和 v2 之间的顺序在这里重要吗？

boost::variant<char*, bool> v1 = "hello";
boost::variant<bool, char*> v2 = "hello";

对于整数，我应该简单地绑定(bind)所有整数的最大整数还是应该单独绑定(bind) int8_t 到 int64_t？如果我将它们全部绑定(bind)，然后输入一个适合它们中的任何一个，它是否会保存为第一个(最后一个？)？

float 和 double 怎么样？

Answer 1

没有魔法。

只有记录在案的构造函数行为。

template<typename T> variant(T & operand);

Requires: T must be unambiguously convertible to one of the bounded types (i.e., T1, T2, etc.).

Postconditions: Content of *this is the best conversion of operand to one of the bounded types, as determined by standard overload resolution rules.

Throws: May fail with any exceptions arising from the conversion of operand to one of the bounded types.

因为这两种情况都涉及隐式转换，所以可能会构造出意想不到的元素类型。

看下面的例子

Live On Coliru

#include <boost/variant.hpp>

int main() {
    {
        boost::variant<bool, std::string> v;
        v = "hello"; // is char const(&)[6], converts to bool

        assert(0 == v.which());

        v = static_cast<char const*>("hello");
        assert(0 == v.which());
    }

    // compare to
    {
        boost::variant<bool, char const*> v;
        v = "hello"; // is char const(&)[6]

        assert(1 == v.which()); // now selects the direct match

        v = static_cast<char const*>("hello");
        assert(1 == v.which());
    }
}