2018-03-26 18:33
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int * time.Second何时工作,什么时候在golang中不工作?

Why does time.Sleep(5 * time.Second) work fine, but:

x := 180
time.Sleep(15 / x * 60 * time.Second)

does not? I get a type mismatch error (types int64 and time.Duration). Given the error, I understand more of why the latter fails than why the former succeeds.

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为什么 time.Sleep(5 * time.Second)可以正常工作,但是:

  x:= 180 
time.Sleep(15 / x * 60 * time.Second)

才不是? 我收到类型不匹配错误(类型为 int64 time.Duration )。 鉴于错误,我更了解后者为何失败而不是前者为何成功。

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  • dsgdfg30210 2018-03-26 18:36

    In Go, a numeric literal (e.g. 60) is an untyped constant. That means it will be silently coerced to whatever type is appropriate for the operation where it's being used. So when you say:

    var x := 5 * time.Second

    Then the type is inferred from time.Second to be a time.Duration, and thus the literal 5 is also treated as a time.Duration. If there's nothing to infer a type from, it will assume a type ("bool, rune, int, float64, complex128 or string") and use that. So:

    x := 180

    Yields x with a type of int.

    However, when you do some operation involving something with a type - like, say a variable x that is an int - then you have two types and one must be converted for the operation to be legal.

    So, to the original question "When does int * time.Second work and when does it not in golang?", int * time.Second actually never works in Go. But 5 * time.Second isn't the same as int * time.Second.

    This is touched on in the Go tour:

    An untyped constant takes the type needed by its context.

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