C++类运算符重载
可以重载类的 not 运算符：
class TestA
{
public:
bool Test;
const bool operator!(){return !Test;}
};
以便...
TestA A; A.Test = false;
if(!TestA){ //etc }
...可以工作。然而，下面的内容是如何工作的呢？
if(TestA) //Does this have to be overloaded too, or is it the not operator inverted?
我要补充一点，读完它后，我对所采用的 typedef 解决方案感到有点困惑，并且我不完全理解正在发生的事情，因为它似乎有些令人困惑。有人可以将其分解为我可以理解的格式吗？
It's possible to overload the not operator for a class:
class TestA
{
public:
bool Test;
const bool operator!(){return !Test;}
};
So that...
TestA A; A.Test = false;
if(!TestA){ //etc }
...can work. However, how does the following work?
if(TestA) //Does this have to be overloaded too, or is it the not operator inverted?
I will add, having read it, I am slightly confused by the typedef solution that is employed, and I don't fully understand what is occurring, as it seems somewhat obsfucated. Could someone break it down into an understandable format for me?
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您可以编写一个
operator bool()。这会处理 bool 的强制转换，使上述语句成为可能。You could write an
operator bool(). This takes care of coercion to bool, making statements as your above one possible.您可以重载
operator void*（如标准库的iostream）或使用boost的技巧，即使用“unspecified_bool_type”typedef（安全布尔）。它不会自动反转您的运算符！You overload
operator void*(like the standard library's iostreams) or use boost's trick of using an "unspecified_bool_type"typedef(safe bool). It does NOT automatically invert youroperator!