c++ - 映射的运算符 [] 无法使用作用域枚举作为键进行编译，但它的 at() 函数有效。为什么会模棱两可？

Question

我的代码如下所示：

enum class MyTypes {
   TYPE_1 = 1,
   TYPE_2 = 2,
   TYPE_3 = 3
};

static const std::regex reg1("some unique expression");
static const std::regex reg2("some unique expression");
static const std::regex reg3("some unique expression");

static const std::map<MyTypes, std::regex> ValidSets = {
    {MyTypes::TYPE_1, reg1},
    {MyTypes::TYPE_2, reg2},
    {MyTypes::TYPE_3, reg3}
};

static const auto match = [&](const std::string& str, const std::regex& reg) -> bool {
    std::smatch base_match;
    return std::regex_match(str, base_match, reg);
};

template<MyTypes TYPE = MyTypes::TYPE_2> // defaulting to a specific type
class Foo {
    std::string sequence_{""};
public:
    const std::uint16_t Type = static_cast<uint16_t>(TYPE);         

    Foo() = default;
    Foo(const std::string_view sequence) {
        assert(!sequence.empty() && "invalid input, must contain at least 1 character.");
        // Fails to compile
        assert(match(sequence.data(), ValidSets[TYPE]) && "invalid sequence set for this type."); 
        // Works
        assert(match(sequence.data(), ValidSets.at(TYPE)) && "invalid sequence set for this type.");
        sequence_ = sequence;
    }
};

然后我将尝试这样使用它：

int main() {
    Foo bar_d("some character sequence"); // must match the regex for the default type

    Foo<Type::TYPE1> bar_1("some character sequence"); // must match the regex for type 1
    Foo<Type::TYPE3> bar_3("some character sequence"); // must match the regex for type 3
 
    return 0;
}

但是，ValidSets[TYPE]在 Visual Studio 2017 编译器错误中无法编译C2678。然而，当我更改它以使用它时，ValidSets.at(TYPE)它编译并执行得很好......

为什么一个编译失败而另一个工作？

标签： c++visual-studio-2017c++17stdmapenum-class