使用mem :: size_of ::< T>时，在编译时无法得知类型为'T'的值的大小.作为数组长度 [英] The size for values of type `T` cannot be known at compilation time when using mem::size_of::<T> as an array length

问题描述

考虑以下功能:

fn make_array<T>()
where
    T: Sized,
{
    let bytes = [0u8; std::mem::size_of::<T>()];
}

无论出于何种原因，它都无法编译

For whatever reason it fails to compile

error[E0277]: the size for values of type `T` cannot be known at compilation time
 --> src/lib.rs:5:23
  |
5 |     let bytes = [0u8; std::mem::size_of::<T>()];
  |                       ^^^^^^^^^^^^^^^^^^^^^^ doesn't have a size known at compile-time
  |
  = help: the trait `std::marker::Sized` is not implemented for `T`
  = note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
  = help: consider adding a `where T: std::marker::Sized` bound
  = note: required by `std::mem::size_of`

尽管事实是通用参数T绑定了Sized特征.对我来说没有任何意义.

This is despite the fact that there is a Sized trait bound for the generic parameter T. It doesn't make any sense to me.

为什么会发生这种情况，我该如何解决?

Why is this happening and how do I work around it?

推荐答案

似乎是错误#43408:数组长度不支持通用参数.

It seems to be bug #43408: Array lengths don't support generic parameters.

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