英文:
Code inside a task expression is executed before the task is created
问题
考虑以下代码：
let makeTask n =
task { return longRunningFunc n }
假设花括号内的所有内容都是任务的一部分，我期望这段代码等同于：
let makeTask n =
Task.Run(fun () -> longRunningFunc n)
但实际上它等同于：
let makeTask n =
let value = longRunningFunc n
Task.Run(fun () -> value)
这既出乎意料，也不太理想，特别是在尝试并行运行多个这样的任务时：
let myTask =
Array.init 100 makeTask // 错误：对 longRunningFunc 的调用是顺序执行的
|> Task.WhenAll
请注意，async 构造器的行为符合预期。我知道任务是“热”的，而 F# 的异步是“冷”的，但我认为这并不能解释这个差异。
我的问题：
- 为什么
task构造器会有这种行为，有没有相关的文档说明？ - 是否有可能让
task构造器按预期将对longRunningFunc的调用移到Task.Run中？
英文:
Consider the following:
let makeTask n =
task { return longRunningFunc n }
Under the assumption that everything inside the curly braces is part of the task, I would expect this to be equivalent to:
let makeTask n =
Task.Run(fun () -> longRunningFunc n)
But it is apparently actually equivalent to:
let makeTask n =
let value = longRunningFunc n
Task.Run(fun () -> value)
Which is both unexpected and far less desirable, for example, when attempting to run many such tasks in parallel:
let myTask =
Array.init 100 makeTask // oops: calls to longRunningFunc occur sequentially
|> Task.WhenAll
Note that the async builder behaves as expected. I understand that tasks are "hot", while F# asyncs are "cold", but I don't think that explains this discrepancy.
My questions:
- Why does the
taskbuilder behave this way, and is it documented anywhere? - Is it possible to get the
taskbuilder to move the call tolongRunningFuncintoTask.Run, as expected?
答案1
得分: 0
F#编译器团队表示这是正确的行为。在任务表达式的顶部执行一个虚拟任务（例如do! Task.Delay(1)）可以修复这个问题，但并不十分优雅。嗯，好吧。
英文:
The F# compiler team says this is the correct behavior. Executing a dummy task at the top of the task expression (e.g. do! Task.Delay(1)) would fix the problem, but isn't very elegant. Oh well.
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