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Which is more preferable to use: lambda functions or nested functions ('def')?
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108
If you need to assign the lambda to a name, use a def instead. defs are just syntactic sugar fo...
What happens to global and static variables in a shared library when it is dynamically linked?
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answered Oct 15 '13 at 6:02
Mikael PerssonMikael Persson
16.3k66 gold badges3333 silver badges4949 bronze badges
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Is Enabling Double Escaping Dangerous?
...ly, a very simple, if limited workaround is simply to avoid '+' and use '%20' instead. In any case, using the '+' symbol to encode a space is not valid url encoding, but specific to a limited set of protocols and probably widely supported for backwards-compatibility reasons. If only for canonicali...
Python - Passing a function into another function
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150
Just pass it in like any other parameter:
def a(x):
return "a(%s)" % (x,)
def b(f,x):
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What does apply_filters(…) actually do in WordPress?
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104
apply_filters($tag, $value) passes the 'value' argument to each of the functions 'hooked' (usin...
Dual emission of constructor symbols
... ::= C3 # complete object allocating constructor
::= D0 # deleting destructor
::= D1 # complete object destructor
::= D2 # base object destructor
Wait, why is this simple? This class has no base. Why does it have a "complete object c...
How do I convert a NSString into a std::string?
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edited May 23 '17 at 10:31
Community♦
111 silver badge
answered Nov 3 '11 at 20:52
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How to load/edit/run/save text files (.py) into an IPython notebook cell?
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answered Jan 10 '14 at 2:00
user2304916user2304916
6,68233 gold badges2727 silver badges4949 bronze badges
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Help with C# generics error - “The type 'T' must be a non-nullable value type”
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Liam
21.3k1717 gold badges9090 silver badges146146 bronze badges
answered Feb 9 '10 at 16:29
Jon SkeetJon Skeet
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When is memoization automatic in GHC Haskell?
...situations. For example, consider the function
f = \x -> let y = [1..30000000] in foldl' (+) 0 (y ++ [x])
GHC might notice that y does not depend on x and rewrite the function to
f = let y = [1..30000000] in \x -> foldl' (+) 0 (y ++ [x])
In this case, the new version is much less effici...