Question

我有一个POD类型Foo我希望用户通过我的用户定义的文字实例化（默认复制，移动和分配都没问题）：

struct Foo
{
private:
    Foo() = delete;

    friend constexpr Foo operator"" _foo (const char* str, std::siz_t n);

    const char* var1;
    std::size_t var2;
};

constexpr Foo operator"" _foo (const char* str, std::siz_t n)
{
    return Foo{str, MyFunc(str, n)};
}

举个例子，我想要这样的东西：

int main()
{
    Foo myBar = "Hello, world!"_foo; // ctor via user defined literal OK
    Foo myBar2 = myBar; // copy-ctor OK
    Foo myBar3 = std::move(myBar2); // move-ctor OK
    myBar = myBar3; // assignment OK

    Foo myBaz{"Hello, world!", 42}; // ERROR: Must go through the user defined literal
    Foo myBaz2; // ERROR: Default ctor not allowed
}

但是使用上面的代码我得到以下编译错误（Xcode 4.6.3 / Clang）：

main.cpp:88:12: error: no matching constructor for initialization of '<anonymous>::Foo'
    return Foo{str, MyFunc(str, n)};
           ^  ~~~~~~~~~~~~~~~~~~~~~
main.cpp:60:5: note: candidate constructor not viable: requires 0 arguments, but 2 were provided
    Foo() = delete;
    ^
main.cpp:57:8: note: candidate constructor (the implicit copy constructor) not viable: requires 1 argument, but 2 were provided
struct Foo
       ^
main.cpp:57:8: note: candidate constructor (the implicit move constructor) not viable: requires 1 argument, but 2 were provided

如何为业务重新打开{} -ctor，但仅针对用户定义的文字？

Answer 1

目前您没有双参数构造函数。 Just write one：

struct Foo
{
private:
    const char* var1;
    std::size_t var2;

    friend constexpr Foo operator"" _foo (const char* v, std::size_t len);

    constexpr Foo(const char* param1, std::size_t param2) : var1(param1), var2(param2) {}
};

如何要求类型通过用户定义的文字？

1 个答案: