如何保证在调用函数对象时参数求值的顺序？ [英] How to guarantee order of argument evaluation when calling a function object?

问题描述

如何避免在使用std :: make_tuple 时避免未定义的构造函数执行顺序的问题的答案导致了在讨论期间，我了解到可以为构造函数保证参数求值的顺序：使用
braced-init-list ，顺序将保证从左到右：

The answers to the question on how to avoid undefined execution order for the constructors when using std::make_tuple led to a discussion during which I learned that the order of argument evaluation can be guaranteed for constructors: Using a braced-init-list the order is guaranteed to be left to right:

T{ a, b, c }

表达式 a ， b 和 c 以给定的顺序评估。这是这种情况，即使 T 只有一个正常的构造函数定义。

The expressions a, b, and c are evaluated in the given order. This is the case, even if the type T just has a normal constructor defined.

显然，是一个构造函数，有时在调用函数时保证求值的顺序是很好的，但是没有诸如 brace-argument-list 这样的函数来调用具有定义的参数求值顺序的函数。问题是：可以使用构造函数的保证来建立一个函数调用设施（ function_apply（））

Clearly, not everything called is a constructor and sometimes it would be nice to guarantee the order of evaluation when calling a function but there is no such thing as brace-argument-list to call function with a defined order of evaluation of their arguments. The question becomes: Can the guarantees to constructors be used to build a function call facility ("function_apply()") with an ordering guarantee for the evaluation of arguments? It is acceptable to require a function object being called.

推荐答案

一个愚蠢的包装类如下：

What about a silly wrapper class like this:

struct OrderedCall
{
    template <typename F, typename ...Args>
    OrderedCall(F && f, Args &&... args)
    {
        std::forward<F>(f)(std::forward<Args>(args)...);
    }
};

用法：

void foo(int, char, bool);

OrderedCall{foo, 5, 'x', false};

如果你想要一个返回值，你可以通过引用传递它以提取返回类型），或将其存储在对象中，以获得如下接口：

If you want a return value, you could pass it in by reference (you'll need some trait to extract the return type), or store it in the object, to get an interface like:

auto x = OrderedCall{foo, 5, 'x', false}.get_result();

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