c – 覆盖常量表达式？

Basically what I’d like to do is build a string class that if handed a static string,or a constexpr variable pointing at one,it will simply use that pointer and won’t later try to delete it.

我想我们这里面临XY-problem.知道表达式是否是constexpr不会告诉您是否适合删除.

我会说如果这个动作有可能不合适,该函数不应该试图猜测是否删除.我相信调用者应该注意这一点,可能使用智能指针来解决这个问题.

换句话说,我会让函数接受左值引用,如果必须将指向对象作为参数传递给函数,则让客户端取消引用指针.

我想说这种模式不仅仅在删除的情况下有意义：更一般地说,如果一段代码有责任决定应该如何生成或计算某些值(例如对象分配),那么还要负责正确执行一些相关或相应的动作(例如清理)应该属于同一段代码：

void foo(my_object& o)
{
    // ...
}

my_object o;
foo(o);
// The argument has automatic storage,no need to delete

my_object* pO = new my_object();
foo(*pO);

// The argument was allocated through operator new,we must
// deallocate it through a corresponding call to delete
delete pO;

如果你真的希望在函数中进行清理,你应该让客户端告诉函数如何执行它：

void foo(my_object& o,bool shouldDelete)
{
    // ...
    if (shouldDelete) { delete &o; }
}

my_object o;
foo(o,false); // "And don't try to deallocate this object please..."

my_object* pO = new my_object();
foo(*pO,true); // "Please,delete it for me" (we are delegating the
                // responsibility of performing the material action,// but not the one of knowing how!)

为了获得更大的灵活性,你甚至可以接受一个可调用的对象,这通过“委派执行物质操作的责任,而不是知道如何”来更清楚地表达我在上述评论中的含义：

#include <functional>

void foo(my_object& o,std::function<void()> f = [] () { })
{
    // ...
    f();
}

int main()
{
    my_object o;
    foo(o); // "And don't do anything with this object when you're done..."

    my_object* pO = new my_object();
    foo(*pO,[=] () { delete pO; }); // "Please,do exactly this when done..."
}

如果您不需要在运行时确定可调用对象的类型,您甚至可以考虑将foo()转换为函数模板.

最后,关于如何确定表达式是否是常量表达式的原始问题,这通常是不可能的,但是有一些技术可以帮助你在cetain情况下 – 只要知道它们的局限性.在这方面,您可能会发现this Q&A on StackOverflow相关.