错误：不能通过`...`传递非trivially可复制类型的对象 [英] error: cannot pass objects of non-trivially-copyable type through `...`

问题描述

我有一个单元，它有

std::is_trivial<unit>::value; // true
std::is_trivially_copyable<unit>::value; // true (on compilers which have this trait)

我想传递的是作为元组，例如

I'd like to pass are vectors of unit as a tuple, e.g.

using geodeticTuple = std::tuple<unit, unit, unit>;

我需要这些向量可以传递到使用不同类型参数的转换函数，例如

I need these vectors to be passable into conversion functions who use different types of arguments, e.g.

someOtherType convert（const geodeticTuple& point，const geodeticTuple& origin）或

someOtherType convert（const geodeticTuple& point，...）

好的，但是用gcc-4.9.3，我得到的错误：

using MSVC2015, this works totally fine, but with gcc-4.9.3, I'm getting the error:

错误：不能传递非trivially类型 const geodeticTuple {aka const struct std :: tuple< unit，unit，unit>} 通过 ...

，因为gcc-4.9.3不支持 is_trivially_xxx 样式类型

and since gcc-4.9.3 doesn't support the is_trivially_xxx style type-traits, I'm having trouble understanding why this is failing.

是一个简单的类型的元组不可以复制吗？

Is a tuple of trivial types not trivially copyable?

推荐答案

tuple 的复制/移动赋值运算符需要对引用类型进行特殊处理，一般情况下。

tuple's copy/move assignment operators require special handling for reference types, so they must be user-provided in the general case.

由于标准不需要微不足道的可复制性，这是一个QoI问题，无论实现者是否提供额外的长度提供保证（这将需要添加额外专业）。

Since the standard doesn't require trivial copyability, it's a QoI issue whether the implementers go to the extra length of providing that guarantee (which would require adding additional specializations).

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