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Do regular expressions from the re module support word boundaries (\b)?
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C++ templates Turing-complete?
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aand now we have concepts lite
– nurettin
Oct 23 '13 at 6:29
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Send an Array with an HTTP Get
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This is indeed unspecified. That's exactly why the answer says "Generally". In strong typed languages the bracket suffixes [] in request parameter names are namely not interpreted the same way as in weak typed languages. It was originally introdu...
What are the downsides to using Dependency Injection? [closed]
...ern here at work and one of our lead developers would like to know: What - if any - are the downsides to using the Dependency Injection pattern?
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Java Equivalent of C# async/await?
...C# developer but occasionally I develop application in Java. I'm wondering if there is any Java equivalent of C# async/await?
In simple words what is the java equivalent of:
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How do I declare a 2d array in C++ using new?
...trayed in the picture. The problem with that statement is it fails to work if M is not constant.
– Mehrdad Afshari
Jun 9 '16 at 23:51
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Any reason why scala does not explicitly support dependent types?
...w that Scala is a lot closer to these other languages than is typically acknowledged.
Despite the terminology, dependent sum types (also known as Sigma types) are simply a pair of values where the type of the second value is dependent on the first value. This is directly representable in Scala,
sc...
How do you represent a JSON array of strings?
...l structure but the string the OP posted is perfectly valid JSON: codebeautify.org/jsonviewer/92ac7b
– ChrisR
May 7 '14 at 9:28
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One DbContext per web request… why?
...simultaneously (which is very common of course). But I expect you already know that and just want to know why not to just inject a new instance (i.e. with a transient lifestyle) of the DbContext into anyone who needs it. (for more information about why a single DbContext -or even on context per thre...
Time complexity of Sieve of Eratosthenes algorithm
...per bound of O(n log log n) + O(n) = O(n log log n) arithmetic operations. If you count bit operations, since you're dealing with numbers up to n, they have about log n bits, which is where the factor of log n comes in, giving O(n log n log log n) bit operations.
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