C++结构体中static constexpr模板函数编译报错原因及解决咨询

最小示例代码 #

#include<array>
#include<iostream>
#include <cstdint>

enum Indices : std::uint16_t
{
    kIndex1 = 0,
    kIndex2 = 1,
    kNumIndices = 2
};

struct TestStruct
{
  static constexpr std::array<uint32_t, Indices::kNumIndices> myvals{4,7};

  template <Indices myIndex>
  static constexpr std::uint32_t GetValue()
  {
    return myvals[myIndex]*5;
  }

  static constexpr std::array<std::uint32_t, Indices::kNumIndices> newVals{GetValue<Indices::kIndex1>(), GetValue<Indices::kIndex2>()};
};

int main()
{
    auto myArr = TestStruct::newVals;
    std::cout<<myArr[Indices::kIndex1]<<"  "<<myArr[Indices::kIndex2]<<std::endl;
}

报错信息 #

Test.cpp:23:97: error: ‘static constexpr uint32_t TestStruct::GetValue() [with Indices myIndex = kIndex1; uint32_t = unsigned int]’ used before its definition
   23 |   static constexpr std::array<uint32_t, Indices::kNumIndices> newVals{GetValue<Indices::kIndex1>(), GetValue<Indices::kIndex2>()};

错误原因 #

这个报错的核心是类内部静态constexpr成员的初始化规则与函数模板可见性的冲突：
虽然GetValue模板函数在newVals前写了完整定义，但C++中类的定义要到闭合的}才算完成。newVals是constexpr静态成员，它的初始化属于常量表达式求值，此时需要实例化GetValue模板，但编译器认为类尚未完整定义，GetValue的定义还未“生效”——常量表达式要求被调用的函数必须已完成定义，因此判定为“使用前未定义”。

不拆分结构体的解决办法 #

办法1：将newVals的初始化移到类外部 #

在类内仅声明newVals，类外完成初始化，此时类已完整定义，GetValue模板可正常实例化：

#include<array>
#include<iostream>
#include <cstdint>

enum Indices : std::uint16_t
{
    kIndex1 = 0,
    kIndex2 = 1,
    kNumIndices = 2
};

struct TestStruct
{
  static constexpr std::array<uint32_t, Indices::kNumIndices> myvals{4,7};

  template <Indices myIndex>
  static constexpr std::uint32_t GetValue()
  {
    return myvals[myIndex]*5;
  }

  // 类内仅声明
  static constexpr std::array<std::uint32_t, Indices::kNumIndices> newVals;
};

// 类外初始化
constexpr std::array<std::uint32_t, Indices::kNumIndices> TestStruct::newVals{GetValue<Indices::kIndex1>(), GetValue<Indices::kIndex2>()};

int main()
{
    auto myArr = TestStruct::newVals;
    std::cout<<myArr[Indices::kIndex1]<<"  "<<myArr[Indices::kIndex2]<<std::endl;
}

办法2：用constexpr lambda延迟求值（C++17及以上） #

利用C++17的constexpr lambda特性，把newVals的初始化逻辑包裹在lambda中，让GetValue的实例化延迟到lambda调用时，此时类已足够完整：

#include<array>
#include<iostream>
#include <cstdint>

enum Indices : std::uint16_t
{
    kIndex1 = 0,
    kIndex2 = 1,
    kNumIndices = 2
};

struct TestStruct
{
  static constexpr std::array<uint32_t, Indices::kNumIndices> myvals{4,7};

  template <Indices myIndex>
  static constexpr std::uint32_t GetValue()
  {
    return myvals[myIndex]*5;
  }

  static constexpr std::array<std::uint32_t, Indices::kNumIndices> newVals = []{
      return std::array<std::uint32_t, Indices::kNumIndices>{
          GetValue<Indices::kIndex1>(), 
          GetValue<Indices::kIndex2>()
      };
  }();
};

int main()
{
    auto myArr = TestStruct::newVals;
    std::cout<<myArr[Indices::kIndex1]<<"  "<<myArr[Indices::kIndex2]<<std::endl;
}

办法3：将GetValue改为变量模板（C++14及以上） #

如果逻辑允许，把函数模板改成变量模板，变量模板的类内求值规则更宽松：

#include<array>
#include<iostream>
#include <cstdint>

enum Indices : std::uint16_t
{
    kIndex1 = 0,
    kIndex2 = 1,
    kNumIndices = 2
};

struct TestStruct
{
  static constexpr std::array<uint32_t, Indices::kNumIndices> myvals{4,7};

  template <Indices myIndex>
  static constexpr std::uint32_t GetValue = myvals[myIndex]*5;

  static constexpr std::array<std::uint32_t, Indices::kNumIndices> newVals{GetValue<Indices::kIndex1>, GetValue<Indices::kIndex2>};
};

int main()
{
    auto myArr = TestStruct::newVals;
    std::cout<<myArr[Indices::kIndex1]<<"  "<<myArr[Indices::kIndex2]<<std::endl;
}

内容的提问来源于stack exchange，提问作者AndreasL