是否可以防止省略聚合初始化成员? [英] Is it possible to prevent omission of aggregate initialization members?

问题描述

我有一个结构，其中有许多相同类型的成员，像这样

I have a struct with many members of the same type, like this

struct VariablePointers {
   VariablePtr active;
   VariablePtr wasactive;
   VariablePtr filename;
};

问题是，如果我忘记初始化一个结构成员(例如 wasactive )，就像这样:

The problem is that if I forget to initialize one of the struct members (e.g. wasactive), like this:

VariablePointers{activePtr, filename}

编译器不会抱怨它，但是我将有一个部分初始化的对象.如何防止这种错误?我可以添加一个构造函数，但是它将重复两次变量列表，因此我必须键入所有这三次！

The compiler will not complain about it, but I will have one object that is partially initialized. How can I prevent this kind of error? I could add a constructor, but it would duplicate the list of variable twice, so I have to type all of this thrice!

如果有针对C ++ 11的解决方案，请也添加 C ++ 11 的答案(当前我仅限于该版本).不过，也欢迎使用最新的语言标准！

Please also add C++11 answers, if there's a solution for C++11 (currently I'm restricted to that version). More recent language standards are welcome too, though!

推荐答案

这是一个技巧，如果缺少所需的初始化程序，则会触发链接器错误:

Here is a trick which triggers a linker error if a required initializer is missing:

struct init_required_t {
    template <class T>
    operator T() const; // Left undefined
} static const init_required;

用法:

struct Foo {
    int bar = init_required;
};

int main() {
    Foo f;
}

结果:

/tmp/ccxwN7Pn.o: In function `Foo::Foo()':
prog.cc:(.text._ZN3FooC2Ev[_ZN3FooC5Ev]+0x12): undefined reference to `init_required_t::operator int<int>() const'
collect2: error: ld returned 1 exit status

注意事项:

• 在C ++ 14之前，这完全阻止了 Foo 的聚合.
• 从技术上讲，这依赖于未定义的行为(违反了ODR)，但应在任何理智的平台上都可以使用.

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