我正在进行一些类似于Y组合器的lambda包装的实验（尽管它们实际上并不严格属于Y组合器，我知道），但是遇到了一个非常奇怪的问题。在Debug配置（关闭优化）下，我的代码运行正常，但是在Release配置（启用 优化（速度为先）(/Ox)）下跳过了很多（而且重要的！）部分。
请注意，lambda函数内部的内容基本上不重要，只是为了确保它可以正确递归等。
// main.cpp
#include <iostream>
#include <string>
#define uint unsigned int
// Defines a y-combinator-style thing to do recursive things. Includes a system where the lambda can declare itself to be obsolete.
// Yes, it's hacky and ugly. Don't worry about it, this is all just testing functionality.
template <class F>
class YCombinator {
public:
F m_f; // the lambda will be stored here
bool m_selfDestructing = false; //!< Whether the combinator will self-destruct should its lambda mark itself as no longer useful.
bool m_selfDestructTrigger = false; //!< Whether the combinator's lambda has marked itself as no longer useful.
// a forwarding operator:
template <class... Args>
decltype(auto) evaluate(Args&&... args) {
// Avoid storing return if we can,
if (!m_selfDestructing) {
// Pass itself to m_f, then the arguments.
return m_f(*this, std::forward<Args>(args)...);
}
else {
// Pass itself to m_f, then the arguments.
auto r = m_f(*this, std::forward<Args>(args)...);
// self-destruct if necessary, allowing lamdas to delete themselves if they know they're no longer useful.
if (m_selfDestructTrigger) {
delete this;
}
return r;
}
}
};
template <class F> YCombinator(F, bool sd)->YCombinator<F>;
// Tests some instances.
int main() {
// Most basic test
auto a = YCombinator{
[](auto & self, uint in)->uint{
uint out = in;
for (uint i = 1u; i < in; ++i) {
out += self.evaluate(i);
}
return out;
},
false
};
// Same as a, but checks it works as a pointer.
auto b = new YCombinator{
[](auto & self, uint in)->uint {
uint out = in;
for (uint i = 0u; i < in; ++i) {
out += self.evaluate(i);
}
return out;
},
false
};
// c elided for simplicity
// Checks the self-deletion mechanism
auto d = new YCombinator{
[&a, b](auto & self, uint in)->uint {
std::cout << "Running d(" << in << ") [SD-" << self.m_selfDestructing << "]..." << std::endl;
uint outA = a.evaluate(in);
uint outB = b->evaluate(in);
if (outA == outB)
std::cout << "d(" << in << ") [SD-" << self.m_selfDestructing << "] confirmed both a and b produced the same output of " << outA << "." << std::endl;
self.m_selfDestructTrigger = true;
return outA;
},
true
};
uint resultA = a.evaluate(4u);
std::cout << "Final result: a(4) = " << resultA << "." << std::endl << std::endl;
uint resultB = (*b).evaluate(5u);
std::cout << "Final result: b(5) = " << resultB << "." << std::endl << std::endl;
uint resultD = d->evaluate(2u);
std::cout << "Final result: d(2) = " << resultD << "." << std::endl << std::endl;
resultD = d->evaluate(2u);
std::cout << "Final result: d(2) = " << resultD << "." << std::endl << std::endl;
}
应该发生的是，第一次对进行评估正常运行，设置了并导致其被删除。然后第二次对进行评估应该崩溃，因为不再真正存在。这正是在Debug配置中发生的情况。(注意：正如@largest_prime_is_463035818在下面指出的那样，它不应该崩溃，而应该遇到未定义的行为。)
但是在Release配置中，据我所知，
evaluate中的所有代码都被完全跳过，执行直接跳转到lambda表达式。显然，在优化后的代码中设置断点有些可疑，但这似乎就是正在发生的事情。我尝试重新构建项目，但是没有成功；VS似乎非常坚定。我疯了吗？我错过了什么吗？或者这是VS（甚至编译器）中的实际错误？任何有助于确定这是否是代码问题或工具问题的帮助都将不胜感激。
注：我使用/ std:c++ latest特性集的VS2019 16.8.3版本。
A;B;而且B没有定义，那么编译器也可以随意处理A。 - 463035818_is_not_a_number