Question

任何人都可以帮忙告诉我为什么参数演绎不符合我的预期吗？请参阅我的代码评论我的思路？

#include <iostream>
#include <type_traits>
#include <iomanip>
#include <string>

using namespace std;

template<class T>
void deduce1(T args, string arg){
cout << "template<class T> void deduce1(T args) " << " argument passed in was: " << arg << " deduced as: " << typeid(T).name() << endl;
cout <<  "Is const: " << boolalpha << is_const<T>::value << endl;
cout <<  "Is reference: " << boolalpha << is_reference<T>::value << endl;
cout <<  "Is pointer: " << boolalpha << is_pointer<T>::value << endl;
}

template<class T>
void deduce2(T& args,string arg){
cout << "template<class T> void deduce2(T args) " << " argument passed in was: " << arg << " deduced as: " << typeid(T).name() << endl;
cout <<  "Is const: " << boolalpha << is_const<T>::value << endl;
cout <<  "Is reference: " << boolalpha << is_reference<T>::value << endl;
cout <<  "Is pointer: " << boolalpha << is_pointer<T>::value << endl;
}

template<class T>
void deduce3(T&& args,string arg){
cout << "template<class T> void deduce3(T args) " << " argument passed in was: " << arg << " deduced as: " << typeid(T).name() << endl;
cout <<  "Is const: " << boolalpha << is_const<T>::value << endl;
cout <<  "Is reference: " << boolalpha << is_reference<T>::value << endl;
cout <<  "Is rvalue reference: " << boolalpha << is_rvalue_reference<T>::value << endl;
cout <<  "Is pointer: " << boolalpha << is_pointer<T>::value << endl;
}

int _tmain(int argc, _TCHAR* argv[])
{

int a = 1;
const int b = 5;
int c[] = {12};
int const d[] = {12};
int& e = a;

deduce1(a,"int a = 1");
deduce1(b,"const int b = 5");
deduce1(c,"int c[] = {12}");
deduce1(d,"int const d[] = {12}"); // would have thought is_const<T> would return true any comments?
deduce1(e,"int& e = a");
deduce1(5,"5");

deduce2(a,"int a = 1");
deduce2(b,"const int b = 5"); //would have though type would be deduced as int const comments?
deduce2(c,"int c[] = {12}"); // why is this not returning true as a reference?
deduce2(d,"int const d[] = {12}"); // would have thought is_const<T> would return true any comments
deduce2(e,"int& e = a");

deduce3(a,"int a = 1");
deduce3(b,"const int b = 5");
deduce3(c,"int c[] = {12}"); // why is this not returning true as a reference?
deduce3(d,"int const d[] = {12}"); // would have thought is_const<T> would return true any comments
deduce3(e,"int& e = a");
deduce3(string("Hello"),"string(\"Hello\")"); // why not rvalue reference


return 0;
}

Answer 1

我将依次处理你的每一条评论。

deduce1与d：
```
template <class T> void deduce1(T args, string arg);
int const d[] = {12};
deduce1(d, "int const d[] = {12}"); // would have thought is_const<T> would return true
```
第一个重要的是你不能通过C ++中的值传递数组作为参数。表示数组的表达式几乎会立即衰减到指向其第一个元素的指针。因此，此处T推导出的类型为int const *。

所以，为什么这不起作用？问题是指针不是const。指针指向的int是const。如果您要报告所指向类型的const，则在检查remove_pointer之前，您需要使用const。因此，您可以将const检查更改为：
```
cout <<  "Is const: " << boolalpha << is_const<typename remove_pointer<T>::type>::value << endl;
```
然而，之前给出的结果是正确答案。指针确实不是const。实际上，即使指针为const，顶级consts也会在类型扣除完成之前被删除，因此T无论如何都不会const。这样做的原因是，如果你通过值传递，你实际上并不关心参数的const，因为你无论如何都要复制它。

deduce2与b：

template <class T> void deduce2(T& args,string arg);
const int b = 5;
deduce2(b,"const int b = 5"); //would have though type would be deduced as int const

不确定这里发生了什么，但我的输出是：

deduced as: i
Is const: true
Is reference: false
Is pointer: false

完全符合预期。

deduce2与c：
```
int c[] = {12};
deduce2(c,"int c[] = {12}"); // why is this not returning true as a reference?
```
它说这不是一个引用，因为所有deduce2调用都说同样的原因（并且deduce3也存在这个问题）。在您的函数中，您正在检查T的类型，但args的类型为T&。您只是在&之前检查零件。所以，args的类型为int (&)[1] c，但您只是在检查int [1]是否为引用。在您的支票中使用decltype(args)代替T进行修复。

deduce2与d：

int const d[] = {12};
deduce2(d,"int const d[] = {12}"); // would have thought is_const<T> would return true

与deduce1 d（第1项）相同的问题。

deduce3与c：

template <class T> void deduce3(T&& args,string arg);
int c[] = {12};
deduce3(c,"int c[] = {12}"); // why is this not returning true as a reference?

deduce2与c（第3项）相同的问题。

deduce3与d：

int const d[] = {12};
deduce3(d,"int const d[] = {12}"); // would have thought is_const<T> would return true

与deduce1 d（第1项）相同的问题。

deduce3与string("Hello")：
```
deduce3(string("Hello"),"string(\"Hello\")"); // why not rvalue reference
```
我假设您对“通用引用”有所了解。 Scott Meyers在这个问题上有一个很棒的article和talk。

在这种情况下，由于string("Hello")是一个右值表达式，因此T类型推导为string（由T&&形式的推导的右值引用类型的规则推导出来）。 args的类型现在为string &&。因此，如果你检查decltype(args)是否是左值，它会说是真的。但T本身并非如此。

如果您修复了代码，以便T上的所有类型特征现在都在decltype(args)，那么您将遇到新问题。当args属于引用类型时，您希望为const的某些类型不会。那是因为没有const引用这样的东西。与在第1项中使用remove_pointer类似，您需要使用remove_reference来获取基础类型并检查它是否为const。

C ++模板参数推导过程

1 个答案: