在lambda表达式中访问constexpr变量，无需捕获

In the following example, I can access the constexpr variable x from inside the lambda y without explicitly capturing it. This is not possible if x is not declared as constexpr.

在下面的示例中，我可以从lambda y中访问constexpr变量x，而无需显式地捕获它。如果x没有声明为constexpr，这是不可能的。

Are there special rules that apply to constexpr for capturing?

对于constexpr是否有适用于捕获的特殊规则?

int foo(auto l) {
    // OK
    constexpr auto x = l();
    auto y = []{return x;};
    return y();

    // NOK
    // auto x2 = l();
    // auto y2 = []{ return x2; };
    // return y2();        
}

auto l2 = []{return 3;};

int main() {
    foo(l2);
}

1 个解决方案

#1

Are there special rules that apply to constexpr for capturing/accessing?

constexpr中是否有用于捕获/访问的特殊规则?

Yes, constexpr variables could be read without capturing in lambda:

是的，constexpr变量可以不用lambda来读取:

A lambda expression can read the value of a variable without capturing it if the variable

lambda表达式可以读取变量的值，而不捕获变量的值

has const non-volatile integral or enumeration type and has been initialized with a constant expression, or

const是否具有非易失性积分或枚举类型，并已用常量表达式初始化

is constexpr and trivially copy constructible.

是可构造的。

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