序列化原始 boost::variant 是否安全?

Is it safe to serialize a raw boost::variant?(序列化原始 boost::variant 是否安全?)

问题描述

boost::variant 声称它是一种值类型.这是否意味着简单地写出 boost::variant 的原始表示并稍后加载它是安全的，只要它只包含 POD 类型?假设它将由相同编译器和相同版本的 boost 编译的代码在相同的架构上重新加载.

boost::variant claims that it is a value type. Does this mean that it's safe to simply write out the raw representation of a boost::variant and load it back later, as long as it only contains POD types? Assume that it will be reloaded by code compiled by the same compiler, and same version of boost, on the same architecture.

另外，(可能)等价地，boost::variant 可以用在共享内存中吗?

Also, (probably) equivalently, can boost::variant be used in shared memory?

推荐答案

关于序列化:它应该有效，是的.但是为什么不使用boost::variant的访问机制来写出variant中包含的实际类型呢?

Regarding serialisation: It should work, yes. But why don't you use boost::variant's visitation mechanism to write out the actual type contained in the variant?

struct variant_serializer : boost::static_visitor<void> {
    template <typename T>
    typename boost::enable_if< boost::is_pod<T>, void>::type
    operator()( const T & t ) const {
        // ... serialize here, e.g.
        std::cout << t;
    }
};

int main() {

    const boost::variant<int,char,float,double> v( '1' );

    variant_serializer s;
    boost::apply_visitor( s, v );

    return 0;
}

关于共享内存:boost::variant 不执行堆分配，所以你可以像 int 一样将它放入共享内存，当然，假设适当的同步.

Regarding shared memory: boost::variant does not perform heap allocations, so you can place it into shared memory just like an int, assuming proper synchronisation, of course.

不用说，正如您所说，以上仅当变体只能包含 POD 类型时才有效.

Needless to say, as you said, the above is only valid if the variant can only contain POD types.

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