问题

// random.h
#pragma once

#include <random>

class Random
{
private:
	static std::uniform_real_distribution<float> sDistribution_;
	static std::mt19937 sGenerator_;
};

// random.cpp
#include "random.h"

std::mt19937 Random::sGenerator_(std::random_device()()); // error here
std::uniform_real_distribution<float> Random::sDistribution_(0.0f, 1.0f);

在编译时出现错误时，可以尝试将 std::random_device() 的括号添加到后面，如上所示。

// random.h
#pragma once

#include <random>

class Random
{
private:
	static std::uniform_real_distribution<float> sDistribution_;
	static std::mt19937 sGenerator_;
};

// random.cpp
#include "random.h"

std::mt19937 Random::sGenerator_(std::random_device()); // error here
std::uniform_real_distribution<float> Random::sDistribution_(0.0f, 1.0f);

答案1

得分: 5

std::mt19937 Random::sGenerator_(std::random_device());

是最令人困扰的解析的一个实例。
您希望执行直接初始化，但实际上您正在：
> 声明Random::sGenerator_为一个函数，该函数接受一个指向不带参数的函数的指针，并返回std::random_device。Random::sGenerator_函数返回std::mt19937。

这是编译器的看法，与Random::sGenerator_的原始定义相冲突，该定义将其声明为类型为std::mt19937的静态数据成员。

解决方案非常简单：std::random_device是一种类型，而不是函数，因此您需要在调用它之前进行初始化，如下所示：

std::mt19937 Random::sGenerator_{std::random_device{}()};
// 或者
std::mt19937 Random::sGenerator_{std::random_device()()};
// 或者
std::mt19937 Random::sGenerator_(std::random_device{}());

通常，一些人认为更喜欢列表初始化是一个好习惯，因为它避免了这种解析模糊性。只需注意，当一个类型具有接受std::initializer_list的构造函数时，它的工作方式会有所不同（对于我们使用的任何类型都不是问题）。

自C++17以来

您还可以在类中初始化静态数据成员，如下所示：

class Random
{
private:
    static inline std::uniform_real_distribution<float> sDistribution_{0, 1};
    static inline std::mt19937 sGenerator_{std::random_device{}()};
};

注意：如果您的类的所有操作都是存储静态数据成员，那么它很可能只应该是一个namespace，而不是一个class。

<details>
<summary>英文:</summary>

```cpp
std::mt19937 Random::sGenerator_(std::random_device());

is an instance of the most vexing parse.
You wanted to perform direct initialization, but you're actually:
> declaring Random::sGenerator_ as a function taking a pointer to a function taking no parameters, and returning std::random_device. The Random::sGenerator_ function returns std::mt19937.

This is what the compiler thinks, and it conflicts with the original definition of Random::sGenerator_, which declares it as a static data member of type std::mt19937.

The solution is very simple: std::random_device is a type, not a function, so you need to initialize it before calling it, like:

std::mt19937 Random::sGenerator_{std::random_device{}()};
// or
std::mt19937 Random::sGenerator_{std::random_device()()};
// or
std::mt19937 Random::sGenerator_(std::random_device{}());

In general, preferring list initialization is considered good practice by some, because it avoids this parsing ambiguity. Just be aware that it works differently when a type has a constructor taking std::initializer_list (not a problem for any of the types we're using).

Since C++17

You can also initialize the static data member in the class, like:

class Random
{
private:
    static inline std::uniform_real_distribution<float> sDistribution_{0, 1};
    static inline std::mt19937 sGenerator_{std::random_device{}()};
};

Note: if all your class does is storing static data members, it should likely just be a namespace, not a class.