在这种情况下如何使用解密?

Here is my code:

这是我的代码:

class Test
{
private:
    SomeType a;
public:
    using TE = decltype(a.find("abc")->second);
    TE getElement(const string &) const;
};

Test::TE Test::getElement(const string & key) const
{
    return a.find(key)->second;
}

In the class, a function can return an element of a like above. The type of a.find(key)->second is very complicated so I don't want to type it all. In fact, I even don't know how to type it...

在类中，函数可以返回与上面类似的元素。查找(关键字)->秒的类型非常复杂，所以我不想把它全部键入。事实上，我甚至不知道如何打字……

So I want to use decltype like above but failed. Here is the error:

所以我想使用上面提到的但失败的解密。这是错误:

error C2227: left of '->second' must point to class/struct/union/generic type

错误C2227: '->秒'左边必须指向类/struct/union/泛型类型

1 个解决方案

#1

You have to either move the definition of a up:

你必须把a的定义往上移动

class Test
{
private:
    SomeType a;
public:
    using T = decltype(a.find("abc")->second);
    ...
};

Or, instead of a, substitute and expression of the correct type using std::declval:

或者用std::代替正确类型的代词和表达式:

class Test
{
public:
    using T = decltype(std::declval<SomeType&>().find("abc")->second);
    ...
private:
    SomeType a;
};

Note that you're missing a > here:

注意，你在这里漏掉了>:

    return a.find(key)-second;
                     ^^^

And that your decltype() expression is looking up a const char[4] instead of a std::string const &. So the most correct version would be:

您的decltype()表达式是查找const char[4]而不是std: string const &。所以最正确的版本应该是:

using T = decltype(std::declval<SomeType&>().find(
                       std::declval<std::string const&>()
                       )->second);

#1