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Comparison between Mockito vs JMockit - why is Mockito voted better than JMockit? [closed]
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Finally, the JMockit Testing Toolkit
has a wider scope and more ambitious
goals than other mocking toolkits, in
order to provide a complete and
sophisticated developer testing
solution. A good API for mocking, even
without artificial limitations, is not
enough for productive cre...
REST API Best practices: Where to put parameters? [closed]
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answered Oct 26 '10 at 23:33
Darrel MillerDarrel Miller
126k2828 gold badges179179 silver badges234234 bronze badges
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Build vs new in Rails 3
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answered Feb 10 '11 at 7:41
henrymhenrym
2,52511 gold badge1616 silver badges1313 bronze badges
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Hyphen, underscore, or camelCase as word delimiter in URIs?
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Neil McGuiganNeil McGuigan
39.6k1010 gold badges100100 silver badges134134 bronze badges
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What part of Hindley-Milner do you not understand?
...ains an element x, whereas Γ ⊢ x : T means that x can be deduced to inhabit type T in context Γ. Considering this, the Var rule reads: »If x is literally contained in the context, it can (trivially) be inferred from it«.
– David
Mar 5 '18 at 15:24
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How Pony (ORM) does its tricks?
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210
Pony ORM author is here.
Pony translates Python generator into SQL query in three steps:
Dec...
Generate a random point within a circle (uniformly)
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Can you explain a bit more how to cut the circle and straighten it out?
– kec
Mar 14 '18 at 17:27
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CMake: How to build external projects and include their targets
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edited May 23 '17 at 12:10
Community♦
111 silver badge
answered Jan 23 '17 at 11:42
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MySQL vs MySQLi when using PHP [closed]
... It's worth noting that things have changed a lot in six years. mysql_*() is now deprecated and will be removed soon. You shouldn't use it for new code.
– user1864610
May 21 '15 at 1:47
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How to find the kth smallest element in the union of two sorted arrays?
...got it, just keep going! And be careful with the indexes...
To simplify a bit I'll assume that N and M are > k, so the complexity here is O(log k), which is O(log N + log M).
Pseudo-code:
i = k/2
j = k - i
step = k/4
while step > 0
if a[i-1] > b[j-1]
i -= step
j += st...