我可以在lambda中使用constexpr值而不捕获它吗？ [英] Can I use a constexpr value in a lambda without capturing it?

问题描述

我想在lambda中使用constexpr值。阅读

请特别注意突出显示的示例。 f（x）不必捕获 x ，因为它没有被滥用（选择过载分辨率） object参数的重载）。同样的论点也适用于您的代码-[basic.def.odr] / 3：

变量 x 其名称显示为可能评估的表达式
ex 由 ex 除非将左值到右值转换
（4.1）应用于 x 会产生一个常数表达式（5.20）不会调用
的任何非平凡函数…

当然可以满足此要求。

...并且，如果 x 是对象，则 ex 是
的元素，是表达式 e 的潜在结果的集合，其中
的左值是右值转换（4.1）应用于 e ，或者 e 是被丢弃的
，值表达式（第5条）。

i 是其潜在结果集按照[basic.def.odr] /（2.1），而ltr转换i

因此，我们已经证明（12.1）不适用-和（ 12.2）显然也不是-Clang拒绝您的代码段是错误的。

I would want to use a constexpr value in a lambda. Reading the answer to Using lambda captured constexpr value as an array dimension, I assumed the following should work:

  #include<array>
  int main()
  { 
    constexpr int i = 0;
    auto f = []{  
      std::array<int, i> a;
    };
    return 0;
  }

However, Clang 3.8 (with std=c++14) complains that

variable 'i' cannot be implicitly captured in a lambda with no capture-default specified

Should this be considered a bug in clang 3.8?

BTW:

The above code does compile with gcc 4.9.2. If I change the lambda expresion to capture explicitly:

...
auto f = [i]{
...

clang 3.8 compiles it, but gcc 4.9.2 fails:

error: the value of ‘i’ is not usable in a constant expression ...

解决方案

Should this be considered a bug in clang 3.8?

Yep. A capture is only needed if [expr.prim.lambda]/12 mandates so:

Note in particular the highlighted example. f(x) does not necessitate x to be captured, because it isn't odr-used (overload resolution selects the overload with the object parameter). The same argumentation applies to your code - [basic.def.odr]/3:

A variable x whose name appears as a potentially-evaluated expression ex is odr-used by ex unless applying the lvalue-to-rvalue conversion (4.1) to x yields a constant expression (5.20) that does not invoke any non-trivial functions…

This requirement is certainly met.

…and, if x is an object, ex is an element of the set of potential results of an expression e, where either the lvalue-to-rvalue conversion (4.1) is applied to e, or e is a discarded-value expression (Clause 5).

i is its set of potential results as per [basic.def.odr]/(2.1), and the l-t-r conversion is indeed immediately applied as its passed to a non-type template parameter of object type.

Hence, as we have shown that (12.1) isn't applicable - and (12.2) clearly isn't, either - Clang is wrong in rejecting your snippet.

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