具有类型规范的boost :: make_optional [英] boost::make_optional with type specification

问题描述

与模板规范一起使用时， boost :: make_optional()行为令我感到困惑.

I'm puzzled about the boost::make_optional() behavior when used with the template specification.

尤其是，我仍然不清楚为什么会这样:

In particular, it's still unclear to me why this:

int pizza = 5;
boost::optional<int> pizza_opt = boost::make_optional<int>(pizza)

引发编译错误无法将类型为'int&&'的右值引用绑定到类型为'int'的左值；而这:

int foo(int bar)
{ return bar; }

boost::optional<int> pizza_opt = boost::make_optional<int>(foo(pizza))

工作正常.

我已经从这得知，使用没有多大意义boost :: make_optional 指定类型，但是我正在阅读一些使用此结构的代码.

I already know from this that it does not make much sense to use boost::make_optional specifying the type, but I'm reading some code which does use of this structure.

谢谢！

推荐答案

boost :: make_optional 的模板参数未在optional内部精确定义类型.

Template parameter of boost::make_optional doesn't define exactly type inside optional.

此模板参数负责完美转发，这是简单的最小示例重现问题:

This template parameter is responsible for perfect forwarding, here is simple minimal example reproducing issue:

#include <iostream>

template<typename T>
void bar(T&& x)
{
    std::cout << __PRETTY_FUNCTION__ << " " 
        << std::forward<T>(x) << '\n';
}

int foo(int x)
{
    return x + 1;
}

int main()
{
    int pizza = 5;
    bar(pizza);
    bar<int>(foo(pizza));
    // bar<int>(pizza); // same error

    return 0;
}

实时演示.

因此，推论完成后，对于L值，T是 int& ；对于r值，T是 int .

So when deduction is done T is int& for l-values and int for r-values.

当您传递变量时，您传递的是左值.

When you pass variable you passing l-value.

指定类型时，您将强制参数为 int&& ，该参数与 int& 不匹配.

When you specified type you are forcing argument to be int && which do not match to int&.

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