Question

在以下Rust汇编者错误中,预期类型与发现类型之间的唯一区别是,有两种寿命,即: 1和 2, 预期,且其发现类型相同, 1和 1。汇编者试图说什么? 如果两生的执行必须相同,则汇编者可以接受<代码> 1为 2。

`impl` item signature doesn t match `trait` item signature
expected signature `fn(& 1 ArrayBase<OwnedRepr<f64>, Dim<[usize; 2]>>) -> std::iter::Map<ndarray::iter::Iter< 2, f64, Dim<[usize; 2]>>, fn(& cellref f64) -> & cellref CellValue< cell>>`
   found signature `fn(& 1 ArrayBase<OwnedRepr<f64>, Dim<[usize; 2]>>) -> std::iter::Map<ndarray::iter::Iter< 1, f64, Dim<[usize; 2]>>, fn(& cellref f64) -> & cellref CellValue< cell>>`

What is the compiler trying to tell you?

Answer 1

这只是错误信息的一部分。通常,该错误的其余部分有说明,解释其意思的是<代码> 1和 2。

In any case, the compiler expects your fn type to return an object with the lifetime 2 but you re giving it an object with a lifetime 1 (which was the reference that was passed in). Usually this means that 2 is a longer lifetime than 1 though it can also mean they are completely unrelated lifetimes or potentially that 1 could live long enough but its used in a type that considers it invariant and thus the compiler is not allowed to satisfy 2.

因此,无论你如何通过<代码>fn。期望这一功能的类型是,根据投入参考,将寿命(或至少不同的寿命)延长的<代码>Iter退回。

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