我正在尝试获取lambda函数的哈希值.为什么我得到两个值(8746164008739和-9223363290690767077)?为什么lambda函数的哈希值不总是一个值?
>>> fn = lambda: 1
>>> hash(fn)
-9223363290690767077
>>> fn = lambda: 1
>>> hash(fn)
8746164008739
>>> fn = lambda: 1
>>> hash(fn)
-9223363290690767077
>>> fn = lambda: 1
>>> hash(fn)
8746164008739
>>> fn = lambda: 1
>>> hash(fn)
-9223363290690767077
Run Code Online (Sandbox Code Playgroud) urllib.request.urlopen
如果response.status_code
不是200,我怎样才能避免例外?现在它提出URLError
或HTTPError
基于请求状态.
有没有其他方法来使用python3基本库进行请求?
如果能获得响应头status_code != 200
?
在一个简单的例子中,通过打印将WHNF转换为NF可以正常工作
Prelude> let x = 1 + 2 :: Int
Prelude> :sprint x
x = _
Prelude> x
3
Prelude> :sprint x
x = 3
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但在某种情况下,类型未声明它不起作用.
Prelude> let x = 1 + 2
Prelude> :sprint x
x = _
Prelude> x
3
Prelude> :sprint x
x = _
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你能解释为什么转换在最后一种情况下不起作用吗?
我可以运行这段代码
fn testf(host: &str) {}
fn start(host: &str) {
testf(host);
testf(host);
}
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但出于某种原因,我无法运行这个:
fn testf(host: &str) {}
fn start(host: &str) {
thread::spawn(move || testf(host));
thread::spawn(move || testf(host));
}
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因为以下错误
src/server.rs:30:5: 30:18 error: the type `[closure@src/server.rs:30:19: 30:38 host:&str]` does not fulfill the required lifetime
src/server.rs:30 thread::spawn(move || testf(host));
^~~~~~~~~~~~~
note: type must outlive the static lifetime
error: aborting due to previous error
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有人可以解释一下,它有什么问题以及如何解决它?
python ×2
closures ×1
exception ×1
hash ×1
haskell ×1
httprequest ×1
lambda ×1
lifetime ×1
python-2.7 ×1
python-3.4 ×1
rust ×1
urllib ×1