[英]Do I have to overload every operator for a class to behave like one of its member variables?
給定用戶定義的類型,如下所示:
struct Word{
std::string word;
Widget widget;
};
有沒有辦法讓這個類的每個重載運算符表現得完全一樣,就像它只是一個字符串一樣? 或者我必須通過以下方式實現該類:
struct Word{
bool operator < (Word const& lhs) const;
bool operator > (Word const& lhs) const;
bool operator <= (Word const& lhs) const;
bool operator => (Word const& lhs) const;
bool operator == (Word const& lhs) const;
bool operator != (Word const& lhs) const;
//etc...
std::string word;
Widget widget;
};
確保我考慮字符串包含的每個重載操作,並將行為應用於字符串值。
我想說你最好的選擇是使用std::rel_ops
,這樣你只需要實現==
和<
並獲得所有這些功能。 這是cppreference的一個簡單示例。
#include <iostream>
#include <utility>
struct Foo {
int n;
};
bool operator==(const Foo& lhs, const Foo& rhs)
{
return lhs.n == rhs.n;
}
bool operator<(const Foo& lhs, const Foo& rhs)
{
return lhs.n < rhs.n;
}
int main()
{
Foo f1 = {1};
Foo f2 = {2};
using namespace std::rel_ops;
std::cout << std::boolalpha;
std::cout << "not equal? : " << (f1 != f2) << '\n';
std::cout << "greater? : " << (f1 > f2) << '\n';
std::cout << "less equal? : " << (f1 <= f2) << '\n';
std::cout << "greater equal? : " << (f1 >= f2) << '\n';
}
如果您需要更完整版本的此類事物,請使用<boost/operators.hpp>
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