c++ - sizeof 是否可以在 lambda 中应用于未捕获的变量，或者这是一个编译器错误？

Question

这是讨论的跟进 found here .

以下代码在 gcc 和 clang ( live demo ) 下编译。对于行 //1 中的情况，这是令人惊讶的，因为 lambda 不捕获任何内容。对于 MCR2 的情况，其中 lambda 返回指针本身，我们得到预期的编译时错误(行 //Will not compile)。运算符 sizeof 的应用与返回指针有何不同？

#include <iostream>

#define MCR1(s) \
  ([]() { return sizeof(s); })()

#define MCR2(s) \
  ([]() { return s; })()

int main() {
  auto *s= "hello world";

  auto x1 = MCR1( s ); //1
  auto y1 = MCR1( "hello world" );
//  auto x2= MCR2( s ); // Will not compile
  auto y2= MCR2( "hello world" );

  std::cout << x1  << "  " << y1  << '\n';
  std::cout // << x2 << "  " 
            << y2 << '\n';
}

---

编辑: 跟进这里的讨论是另一个例子。令人惊讶的是，标记为 //2 的行现在可以在 gcc7(开发版)下编译 ( live demo )。此处的区别在于表达式现在标记为 constexpr。

#include <iostream>

#define MCR1(s) \
  ([]() { return sizeof(s); })()

#define MCR2(s) \
  ([]() { return s; })()

int main() {
  auto constexpr *s= "hello world";

  auto constexpr x1= MCR1( s );
  auto constexpr y1= MCR1( "hello world" );
  auto constexpr x2= MCR2( s );             //2
  auto constexpr y2= MCR2( "hello world" );

  std::cout << x1 << "  " << y1 << '\n';
  std::cout << x2 << "  " << y2 << '\n';
}

Answer 1

不同之处在于(缺乏)上下文评估。 sizeof 未计算。

根据 N3337(≈C++11)

§5.1 2 [expr.prim.lambda] / 11

If a lambda-expression has an associated capture-default and its compound-statement odr-uses this or a variable with automatic storage duration and the odr-used entity is not explicitly captured, then the odr-used entity is said to be implicitly captured;

和

§5.1.2 [expr.prim.lambda] / 12

If a lambda-expression odr-uses this or a variable with automatic storage duration from its reaching scope, that entity shall be captured by the lambda-expression. If a lambda-expression captures an entity and that entity is not defined or captured in the immediately enclosing lambda expression or function, the program is ill-formed.

ODR 使用意味着在潜在评估的上下文中使用:

§3.2 [basic.def.odr] / 2

An expression is potentially evaluated unless it is an unevaluated operand or a subexpression thereof. A variable whose name appears as a potentially-evaluated expression is odr-used unless it is an object that satisfies the requirements for appearing in a constant expression and the lvalue-to-rvalue conversion is immediately applied

因为 sizeof 不是，并且 s 在 lambda 表达式的 reaching scope 中，所以没关系。不过，返回 s 意味着对其求值，这就是它格式错误的原因。