c++ - C++模板实参推演过程

Question

任何人都可以帮助告诉我为什么论点推论没有按我的预期工作吗？请查看我的代码注释以了解我的思路？

#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;
}

score 0 · Accepted Answer

我会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
```
第一件重要的事情是，您不能在 C++ 中按值将数组作为参数传递。表示数组的表达式几乎会立即衰减为指向其第一个元素的指针。所以T这里推导出的类型将是int const *.

那么，为什么这不起作用？问题是指针不是const. 指针指向的ints 是const。如果要报告const所指向类型的正确性，则需要remove_pointer在检查const正确性之前使用。因此，您将const检查更改为：
```
cout <<  "Is const: " << boolalpha << is_const<typename remove_pointer<T>::type>::value << endl;
```
但是，它之前给出的结果是正确的答案。指针确实不是const。事实上，即使指针是const，顶层consts在类型推导完成之前被剥离，所以无论如何T都不会const。这样做的原因是，如果您按值传递，那么您实际上并不关心const参数的有效性，因为无论如何您都在复制它。

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?
```
它是说它不是一个参考，出于同样的原因，所有的deduce2电话都这么说（你也有这个问题deduce3）。在您的函数中，您正在检查类型，T但类型args是T&。您只检查&. 所以在这里，argsis的类型，int (&)[1] c但您只是在检查是否int [1]是参考。通过在检查中使用decltype(args)而不是修复它。T

deduce2与d：

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

deduce1与（第 1 项）相同的问题d。

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（第 3 项）相同的问题c。

deduce3与d：

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

deduce1与（第 1 项）相同的问题d。

deduce3与string("Hello")：
```
deduce3(string("Hello"),"string(\"Hello\")"); // why not rvalue reference
```
我将假设您对“通用参考”有所了解。Scott Meyers 有一篇很棒的文章和关于这个主题的演讲。

在这种情况下，因为string("Hello")是一个右值表达式，所以类型T被推导为string（通过推导右值引用类型的规则T&&）。的类型args是现在string &&。所以如果你检查是否decltype(args)是一个右值，它会说是真的。但T它本身不是。

如果您修复代码以使所有类型特征 onT现在都 on decltype(args)，您将遇到一个新问题。当 args 是引用类型时，您期望的某些类型const不会是。那是因为没有const参考之类的东西。与第 1 项中的使用类似remove_pointer，您需要使用remove_reference来获取底层类型并检查是否为const.

c++ - C++模板实参推演过程

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