访问概括捕获在拉姆达对象后返回类型

Question

int main() 
{ 
    auto l = [x = 10]() -> decltype(x) {}; 
} 

• clang++ 4.0 rejects this code，错误如下：

  error: use of undeclared identifier 'x' 
       auto l = [x = 10]() -> decltype(x) {}; 
               ^
  

• g++ 7 rejects this code，错误如下：

  In function 'int main()': 
  error: 'x' was not declared in this scope 
      auto l = [x = 10]() -> decltype(x) {}; 
             ^
  error: 'x' was not declared in this scope 
  In lambda function: 
  warning: no return statement in function returning non-void [-Wreturn-type] 
      auto l = [x = 10]() -> decltype(x) {}; 
             ^
  

这是一个错误还是标准中有某些东西可以显式地阻止在lambda的尾随返回类型中使用用C++ 14通用语法捕获的对象？

---

注意，两种编译器很高兴与非广义捕获：

int main() 
{ 
    int x = 10; 
    auto l = [x]() -> decltype(x) { return 0; }; 
} 

Answer 1

TL; DR：编译器像预期的那样。

该标准定义的λ语义如下[expr.prim.lambda，部分1]：

lambda-expression:

lambda-introducer lambda-declarator_opt compound-statement 

这里化合物语句之间{}拉姆达的只是身体，因为其他一切被包括在拉姆达说明符：

lambda-declarator:

(parameter-declaration-clause) decl-specifier-seq_opt 
     exception-specification_opt attribute-specifier-seq_opt trailing-return-type_opt 

此外，在同一章的第12条，它说，

An init-capture behaves as if it declares and explicitly captures a variable of the form “auto init-capture ;” whose declarative region is the lambda-expression’s compound-statement, except that:

(12.1) — if the capture is by copy (see below), the non-static data member declared for the capture and the variable are treated as two different ways of referring to the same object, which has the lifetime of the non-static data member, and no additional copy and destruction is performed, and

(12.2) — if the capture is by reference, the variable’s lifetime ends when the closure object’s lifetime ends.

因此，在您的第一个示例中，变量x作用域仅为lambda体，不包括decltype表达式。在第二个例子中，显然，x范围是功能main。