小编Tro*_*iar的帖子

问题:是否有可能通过将其参数传递给内部constexpr函数(可能带有某种"完美转发")来评估函数内部的常量表达式？例:

constexpr size_t foo(char const* string_literal) {
    return /*some valid recursive black magic*/;
}

void bar(char const* string_literal) {
    // works fine
    constexpr auto a = foo("Definitely string literal.");
    // compile error: "string_literal" is not a constant expression
    constexpr auto b = foo(string_literal);
}

template<typename T>
void baz(T&& string_literal) {
    // doesn't compile as well with the same error
    constexpr auto b = foo(std::forward<T>(string_literal));
}

int main() {
    // gonna do this, wont compile due to errors mentioned above
    bar("Definitely …

有没有办法创建基类(如boost :: noncopyable)并从中继承,如果它不是由用户(开发人员)制作的,它将禁止编译器为派生类生成默认构造函数？

例:

class SuperDad {
XXX:
  SuperDad(); // = delete?
};

class Child : YYY SuperDad {
public:
  Child(int a) {...}
};

结果:

int main () {
  Child a;     // compile error
  Child b[7];  // compile error
  Child c(13); // OK
}