在没有指定捕获默认值的模板 lambda 中隐式捕获 const 变量

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？这是 Clang 尚未实现的某些编译器扩展还是只是 Clang 错误？

Answer 1

是的，Clang 错误。

适用的规则来自[basic.def.odr]/9:

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

值得注意的是，规则说的是“如果您使用它，它的格式不正确”，而不是“如果我们不能确定它是否是一个 odr-使用，它的格式不正确”。

在该函数调用运算符模板的任何特化中都没有对 x 的 ODR 使用。

Answer 2

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.

在我们的示例中：

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

x 在 lambda 主体中不可用，因为中间范围不捕获 x（既不是简单捕获也不是捕获默认值）。

因此，它不可用。但它是 odr-used 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（为了方便，我们的变量甚至命名为 x！）x 可用于常量表达式，因为它是常量整数类型，而 E 是左值-此处为右值转换。

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
  };
}

在没有指定捕获默认值的模板 lambda 中隐式捕获 const 变量

Implicitly capture const variable in a template lambda with no capture-default specified

c++

lambda

language-lawyer

c++17

c++20