有没有更安全的方法来使用联合在整数和浮点数之间进行转换? [英] Is there a safer way to use unions to convert between integer and floating-point numbers?

问题描述

我正在用 Rust 编写一个 VM，我有 C 和 C++ 背景.我需要类似联合的功能，因为在 VM 堆栈上，我可以存储 int 或 float.

I´m writing a VM in Rust and I have a C and C++ background. I need union-like functionality because on the VM stack I can either store an int or a float.

在 C 中我有一个联合:

In C I had a union:

union stack_record_t {
    int i;
    float f;
};

我可以将记录作为 int 或 float 使用，运行时开销为零.我有一个静态字节码分析器，它会在字节码执行之前发现类型错误，所以我不必在记录旁边存储标志.

I can use the record as int or as float with zero runtime overhead. I have a static bytecode analyzer which will find type errors before the bytecode executes, so I don't have to store a flag alongside the record.

我不知道在 Rust 中使用联合是否是个好主意，因为它们不安全.在 Rust 中有什么安全的方法可以做到这一点 - 也是零成本吗?我应该只使用不安全的 Rust 联合吗?

I don´t know if it is a good idea to use unions in Rust because they are unsafe. Is there any safe way to do this in Rust - also with zero cost? Should I just use the unsafe Rust unions?

推荐答案

您可以使用 f32::from_bits 和 to_bits 安全地将 u32 的原始位重新解释为 f32，反之亦然.这是一个免费"的转换–它编译为无代码(打开优化).¹要在 u32 和 i32 之间进行转换，您可以使用 as 强制转换，这在用于更改签名时同样是免费的.

You can use f32::from_bits and to_bits to safely reinterpret the raw bits of a u32 as an f32 and vice versa. This is a "free" conversion – it compiles to no code (with optimizations turned on).¹ To convert between u32 and i32 you can use as casts, which are likewise free when used to change signedness.

在我看来 u32 是这里的公分母，因此您可以考虑制作一个包含 u32 并公开方法的 struct获取或设置适当的类型:

It seems to me that u32 is the common denominator here, so you might consider making a struct that contains a u32 and exposes methods to get or set the appropriate type:

pub struct Record(u32);

impl Record {
    fn get_int(&self) -> i32 {
        self.0 as _
    }

    fn get_float(&self) -> f32 {
        f32::from_bits(self.0)
    }

    fn set_int(&mut self, value: i32) {
        self.0 = value as _;
    }

    fn set_float(&mut self, value: f32) {
        self.0 = value.to_bits();
    }
}

比较生成的代码.

• 是否可以写Quake的快Rust 中的 InvSqrt() 函数?
• 正在转换字节浮动安全或它可能会产生未定义的行为?
• 整数之间的转换是否昂贵?

¹这些函数在内部使用 transmute，就像使用联合一样重新解释位.所以当它们被优化器内联时，生成的代码是相同的.

¹ These functions use transmute internally, which reinterprets the bits just as using a union would. So when they are inlined by the optimizer the generated code is the same.

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