了解关于左值到右值转换的示例

问题描述

限时删除！！

我很难理解这段代码（来自C ++ 14草案标准 [conv.lval] 的示例）如何调用未定义的行为 g（false）。为什么 constexpr 使程序有效？

此外，不访问 yn ？在 g（）的两个调用中，我们返回了 n 数据成员为什么最后一行说' t访问它？

  struct S {int n; }; 
 auto f（）{
 S x {1}; 
 constexpr S y {2}; 
 return [&]（bool b）{return（b？y：x）.n; }; 
} 
 auto g = f（）; 
 int m = g（false）; //未定义的行为，由于访问x.n外面的
 //生命周期
 int n = g（true）; // OK，does not access yn 
  

解决方案

yn 不是odr使用，因此不需要访问 yn odr使用的规则覆盖 3.2 并说：

注意，Ben Voigt做了一些有用的评论，这一点。因此，这里的工作假设是 x ：

  y 
  

和 e 将是（ e定义的不同表达式在第2 ）：

$ c>

y 产生常数表达式，并将左值到右值转换应用于

由于 f 产生一个lambda，捕获<$ c $一旦调用 f ，引用 x 的c> f 因为 x 是 f 中的自动变量。由于 y 是一个常量表达式，它的行为好像 yn

您的示例包含在 section 4.1 [conv.lval] 说：

并包括考试所属的以下项目符号：

然后：

由于。

I have a hard time understanding how this code (an example from the C++14 draft standard [conv.lval]) invokes undefined behavior for g(false). Why does constexpr make the program valid?

Also, what does it mean by "does not access y.n"? In both calls to g() we are returning the n data member so why does the last line say it doesn't access it?

struct S { int n; };
auto f() {
    S x { 1 };
    constexpr S y { 2 };
    return [&](bool b) { return (b ? y : x).n; };
}
auto g = f();
int m = g(false); // undefined behavior due to access of x.n outside its
                  // lifetime
int n = g(true);  // OK, does not access y.n

解决方案

This is because y.n is not odr-used and therefore does not require an access to y.n the rules for odr-use are covered in 3.2 and says:

Note, Ben Voigt made some helpful comments that clarified this one a bit. So the working assumption here is that x would be:

y

and e would be(the different expression that e is defined for is covered in paragraph 2 of section 3.2):

(b ? y : x).n

y yields a constant expression and the lvalue-to-rvalue conversion is applied to the expression e.

Since f yields a lambda which captures f's local variables by reference x is no longer valid once the call to f is done since x is an automatic variable inside f. Since y is a constant expression it acts as if y.n was not accessed and therefore we don't have the same lifetime issue.

Your example is included in N3939 section 4.1 [conv.lval] and right before that example it says:

and includes the following bullet which the examle belongs to:

then:

This was applied to the C++14 draft standard due to defect report 1773 .

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