运算符重载中的隐式构造？
是否可以使用带有运算符的显式构造？
就像这个例子一样（这当然不起作用）：
class myFoo {
public:
double x, y;
myFoo(double, double);
void operator [] (myFoo);
};
int main() {
myFoo f1(0.0, 1.1);
f1[ {9.9, 10.0} ]; /// or whatever syntax to use, does not work
f1.operator[] ( {9.9, 10.0} ); /// works !
}
Is it possible use in-explicit constructing with operators ?
Just like in this example (which does of course not work):
class myFoo {
public:
double x, y;
myFoo(double, double);
void operator [] (myFoo);
};
int main() {
myFoo f1(0.0, 1.1);
f1[ {9.9, 10.0} ]; /// or whatever syntax to use, does not work
f1.operator[] ( {9.9, 10.0} ); /// works !
}
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评论（5）
C++11 允许将初始值设定项列表传递给重载的
运算符[]（请参阅 13.5.5）。您需要类似的内容
，这将与
f1[ {9.9, 10.0} ];的原始语法相匹配。C++11 allows an initializer list to be passed to an overloaded
operator [](see 13.5.5).You'd need something like
That would match your original syntax of
f1[ {9.9, 10.0} ];.嗯...
f1[myFoo(9.9, 10.0)];？需要一个合理的
operator[]（即期望一个const myFoo&绑定到临时变量）。hmm...
f1[myFoo(9.9, 10.0)];?requires a reasonably sane
operator[](i.e. expects aconst myFoo&to bind to the temporary).不，在当前的 C++03 或即将推出的 C++11 标准中使用
operator []是不可能的。但是，您可以使用一些奇怪的语法来利用operator []来实现相同的目的：其中
operator []如下：No this is not possible using
operator []in current C++03 or upcoming C++11 standard. However, you can exploitoperator []with some weird syntax to achieve the same:where the
operator []is as,是的，它确实适用于当前的代码。无论您使用“[]”实现什么（我所做的就是返回类内数组中索引处的值。因此该类现在可以用作数组类型）运算符所有需要完成的操作确保您传入 const 引用并且不进行任何更改。
会工作的。
yes it does work with the current code. Whatever might be your implementation with the '[]' (all I have done is to return the value at the index in an array inside a class. So the class can now be used as an array type) operator all that needs to be done to make sure that you pass in a const reference and make no changes are done.
Will work.
您的代码在 C++11（g++ 版本 4.5.2）中运行良好。你只需说
而不是
Your code works fine in C++11 (g++ Version 4.5.2). You just have to say
instead of