为什么sizeof（std :: variant）的大小与具有相同成员的结构的大小相同？ [英] Why is sizeof(std::variant) the same size as a struct with the same members?

问题描述

类模板 std :: variant 表示类型安全的联合。 std :: variant 的实例在任何给定时间要么具有其替代类型之一的值，要么不具有任何值。

The class template std::variant represents a type-safe union. An instance of std::variant at any given time either holds a value of one of its alternative types, or it holds no value.

sizeof(std::variant<float, int32_t, double>) == 16

但是如果是工会，为什么要占用这么多空间？

But if it is a union, why does it take so much space?

struct T1 {
    float a;
    int32_t b;
    double c;
};

struct T2 {
    union {
        float a;
        int32_t b;
        double c;
    };
};

变量的大小与结构相同

sizeof(T1) == 16
sizeof(T2) == 8

我希望联合的大小加上要存储的4个字节（该类型是活动的）。

I would expect the size of the union, plus 4 bytes to store, which type is active.

推荐答案

此处，在变量中具有最大对齐方式的类型是 double ，对齐方式为8。这意味着整个 variant 必须对齐方式为8。这也意味着其大小必须为8的倍数，以确保数组中每个 variant 的 double 均应对齐8。

Here the type with the largest alignment in the variant is double, with an alignment of 8. This means the entire variant must have an alignment of 8. This also means its size must be a multiple of 8, ensuring that in an array each variant will have its double aligned by 8.

但是一个变体不仅要存储最大的类型，还必须存储一个标识符或标记，以了解当前实例化的类型。因此，即使仅将1个字节用于该标识符，整个结构也会填充为16，因此它是8的倍数。

But a variant must store more than just the largest type: it must also store an identifier or tag to know which type is currently instantiated. So even if only 1 byte is used for that identifier, the structure as a whole is padded to 16 so that it is a multiple of 8 in size.

更正确比较如下：

struct
{
    union
    {
        float a;
        int32_t b;
        double c;
    };
    int identifier;
};

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