"If constexpr"在C ++ 17中无法在非模板函数中工作 [英] "If constexpr" in C++17 does not work in a non-templated function

问题描述

我尝试使用C ++ 17标准.我试图使用C ++ 17 if constexpr的功能之一.我有一个问题...请看下面的代码.编译没有错误.在下面的代码中，我尝试使用if constexpr检查它是否是指针.

I tried to play with the C++17 standard. I tried to use one of the features of C++17 if constexpr. And I had a problem... Please take a look at the following code. This compiles without errors. In the following code, I tried to use if constexpr to check if it is a pointer.

#include <iostream>
#include <type_traits>

template <typename T>
void print(T value)
{
  if constexpr (std::is_pointer_v<decltype(value)>)
    std::cout << "Ptr to " << *value << std::endl; // Ok
  else
    std::cout << "Ref to " << value << std::endl;
}

int main()
{
  auto n = 1000;
  print(n);
  print(&n);
}

但是当我重写上面的代码时，如下所示，其中if constexpr在main函数中:

But when I rewrite the above code, as shown below, where if constexpr is in the main function:

#include <iostream>
#include <type_traits>

int main()
{
  auto value = 100;
  if constexpr (std::is_pointer_v<decltype(value)>)
    std::cout << "Ptr to " << *value << std::endl; // Error
  else
    std::cout << "Ref to " << value << std::endl;
}

我收到编译错误:

main.cpp:8:32: error: invalid type argument of unary ‘*’ (have ‘int’) 
std::cout << "Ptr to " << *value << std::endl;

问题不在主要功能中.该功能可以是类似于以下功能的任何功能.

Problem is not in the main function. This can be any function similar to the following.

void print()
{
  auto value = 100;
  if constexpr (std::is_pointer_v<decltype(value)>)
    std::cout << "Ptr to " << *value << std::endl; // Error
  else
    std::cout << "Ref to " << value << std::endl;
}

int main()
{
  print();
}

我想知道为什么if constexpr仅在模板函数中起作用，即使类型是由输入参数的decltype推导出来的.

I would like to know why if constexpr works only in template functions, even if the type is deduced by the decltype from the input parameter.

推荐答案

我想知道为什么"if constexpr"仅在模板函数中起作用，即使类型是由decltype从输入参数中推导出来的.

I would like to know why "if constexpr" works only in template functions, even if the type is deduced by the decltype from the input parameter.

这是设计使然.

if constexpr在 template 内，则不会实例化未采用的分支.它不仅会将不被视为标记的分支视为令牌，而且避免对其进行解析或完全执行语义分析.双方仍将进行分析，并且由于*value对于int而言格式不正确，所以这是一个错误.

if constexpr will not instantiate the branch not taken if it's within a template. It won't just treat the branch not taken as token soup and avoid parsing it or performing semantic analysis entirely. Both sides are still going to be analyzed, and since *value is ill-formed for ints, that's an error.

您根本无法使用if constexpr来避免编译非模板代码.只是避免实例化对于特定专业化可能无效的模板代码.

You simply can't use if constexpr to avoid compiling non-template code. It's only to avoid instantiating template code that's potentially invalid-for-the-particular-specialization.

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