c++ - 在模板 lambda 中隐式捕获 const 变量，而未指定捕获默认值

Question

考虑以下代码:

auto f() {
  const auto x = 1;
  return [] (auto) { return x; };
}

GCC and MSVC compiles fine but Clang rejected it .我应该信任哪个编译器？那是 Clang 尚未实现的编译器扩展还是只是 Clang 错误？

Answer 1

这是一个 clang 错误。
我们的规则是 [basic.def.odr]/9 :

A local entity is odr-usable in a scope if:

• either the local entity is not *this, or an enclosing class or non-lambda function parameter scope exists and, if the innermost such scope is a function parameter scope, it corresponds to a non-static member function, and
• for each intervening scope ([basic.scope.scope]) between the point at which the entity is introduced and the scope (where *this is considered to be introduced within the innermost enclosing class or non-lambda function definition scope), either:
  • the intervening scope is a block scope, or
  • the intervening scope is the function parameter scope of a lambda-expression that has a simple-capture naming the entity or has a capture-default, and the block scope of the lambda-expression is also an intervening scope.

If a local entity is odr-used in a scope in which it is not odr-usable, the program is ill-formed.

在我们的例子中:

auto f() {
  const auto x = 1;
  return [] (auto) { return x; };
}

x由于不捕获 x 的介入范围，因此无法在 lambda 主体中使用 odr (既不是简单捕获也不是默认捕获)。
所以，它不是 odr 可用的。但它是 odr 使用的吗？不，来自 [basic.def.odr]/4 :

A variable is named by an expression if the expression is an id-expression that denotes it. A variable x whose name appears as a potentially-evaluated expression E is odr-used by E unless

• x is a reference that is usable in constant expressions ([expr.const]), or
• x is a variable of non-reference type that is usable in constant expressions and has no mutable subobjects, and E is an element of the set of potential results of an expression of non-volatile-qualified non-class type to which the lvalue-to-rvalue conversion ([conv.lval]) is applied, or
• x is a variable of non-reference type, and E is an element of the set of potential results of a discarded-value expression ([expr.prop]) to which the lvalue-to-rvalue conversion is not applied.

第二个要点适用(我们的变量甚至被命名为 x !) x可用于常量表达式，因为它是常量整数类型和 E这里是左值到右值的转换。
所以 x不是 odr-usable，也不是 odr-used，所以这里没有问题。

事实上，我们甚至在 [expr.prim.lambda.capture]/7 中有这个例子。 :

void f(int, const int (&)[2] = {});         // #1
void f(const int&, const int (&)[1]);       // #2
void test() {
  const int x = 17;
  auto g = [](auto a) {
    f(x);                       // OK: calls #1, does not capture x
  };
}