为什么 g++ 发出（而不是发出）-Wreturn 类型警告

Question

#include <array>\n#include <iostream>\n#include <string>\nusing namespace std;\n\ntemplate <typename T> class MyClass {\npublic:\n  // even not called, g() has -Wreturn-type warning\n  double g() { std::cout << "g() called" << std::endl; }\n  // only that f() is called, f() has -Wreturn-type warning otherwise, there is\n  // no -Wreturn-type warning for it\n  MyClass &f() { std::cout << "f() called" << std::endl; }\n  // even not called and not instantiated, h() has -Wreturn-type warning\n  // ! Interestingly, if i change return type from double to MyClass, there is\n  // no -Wreturn-type warning if not called!!! Why? what the compiler is doing\n  // under the hood?\n  template <typename U> double h() {\n    std::cout << "h() called" << sizeof(U) << std::endl;\n  }\n};\n\nint main() {\n  MyClass<double> m;\n  // if f() is not called, there will be no -Wreturn-type warning\n  // m.f();\n\n  (void)m;\n}\n

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当我使用以下命令进行编译时：

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#include <array>\n#include <iostream>\n#include <string>\nusing namespace std;\n\ntemplate <typename T> class MyClass {\npublic:\n  // even not called, g() has -Wreturn-type warning\n  double g() { std::cout << "g() called" << std::endl; }\n  // only that f() is called, f() has -Wreturn-type warning otherwise, there is\n  // no -Wreturn-type warning for it\n  MyClass &f() { std::cout << "f() called" << std::endl; }\n  // even not called and not instantiated, h() has -Wreturn-type warning\n  // ! Interestingly, if i change return type from double to MyClass, there is\n  // no -Wreturn-type warning if not called!!! Why? what the compiler is doing\n  // under the hood?\n  template <typename U> double h() {\n    std::cout << "h() called" << sizeof(U) << std::endl;\n  }\n};\n\nint main() {\n  MyClass<double> m;\n  // if f() is not called, there will be no -Wreturn-type warning\n  // m.f();\n\n  (void)m;\n}\n

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编译器问题：

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app.cpp: In member function \xe2\x80\x98double MyClass<T>::g()\xe2\x80\x99:\napp.cpp:10:56: warning: no return statement in function returning non-void [-Wreturn-type]\n   double g() { std::cout << "g() called" << std::endl; }\n                                                        ^\napp.cpp: In member function \xe2\x80\x98double MyClass<T>::h()\xe2\x80\x99:\napp.cpp:20:3: warning: no return statement in function returning non-void [-Wreturn-type]\n   }\n

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在我的所有函数中，我都没有故意返回，\n我不使用 function f()，因此编译器不会发出警告，可能会被优化掉，这是可以理解的；但我都不用h()，为什么会有警告？

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更奇怪的是，如果我将h()from的返回类型更改double为 to MyClass，警告就会消失

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那么编译器会看到这些函数什么以及在编译过程中如何处理它们呢？

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Godbolt 演示

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Answer 1

依赖返回类型可以（有时）是void：

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template<class T>\nstruct X {\n  std::conditional_t<sizeof(T)<17,void,int> f() {}\n};\n

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GCC 只是等待检查警告，直到知道类型为止。It\xe2\x80\x99s \xe2\x80\x9cobviously\xe2\x80\x9d 从未出现void在您的示例中，但请考虑条件可能是任意 constexpr 计算。

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