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Why are functions in Ocaml/F# not recursive by default?
...uence in OCaml:
let shannon fold p =
let p x = p x *. log(p x) /. log 2.0 in
let p t x = t +. p x in
-. fold p 0.0
Note how the argument p to the higher-order shannon function is superceded by another p in the first line of the body and then another p in the second line of the body.
Conver...
JavaScript for…in vs for
...which idiom is best understood.
An array is iterated using:
for (var i = 0; i < a.length; i++)
//do stuff with a[i]
An object being used as an associative array is iterated using:
for (var key in o)
//do stuff with o[key]
Unless you have earth shattering reasons, stick to the establis...
Is there a performance difference between i++ and ++i in C?
...
406
Executive summary: No.
i++ could potentially be slower than ++i, since the old value of i
mig...
Chaining multiple MapReduce jobs in Hadoop
...
answered Mar 24 '10 at 22:31
Binary NerdBinary Nerd
13.1k44 gold badges3737 silver badges4141 bronze badges
...
Saving and Reading Bitmaps/Images from Internal memory in Android
... OutputStream
bitmapImage.compress(Bitmap.CompressFormat.PNG, 100, fos);
} catch (Exception e) {
e.printStackTrace();
} finally {
try {
fos.close();
} catch (IOException e) {
e.printStackTrace();
...
Can a Byte[] Array be written to a file in C#?
... |
edited Jul 17 '13 at 0:17
answered Dec 19 '08 at 16:58
...
How to launch html using Chrome at “--allow-file-access-from-files” mode?
... |
edited Mar 17 '14 at 10:47
Saran
3,67133 gold badges3232 silver badges5353 bronze badges
answered Se...
Why do we need a pure virtual destructor in C++?
...unctions can have implementations).
struct foo {
virtual void bar() = 0;
};
void foo::bar() { /* default implementation */ }
class foof : public foo {
void bar() { foo::bar(); } // have to explicitly call default implementation.
};
...
Convert string with comma to integer
...
Michael KohlMichael Kohl
62k1010 gold badges125125 silver badges149149 bronze badges
...
What exactly is a reentrant function?
...t, everything seems ok… But wait:
int main()
{
foo(bar);
return 0;
}
If the lock on mutex is not recursive, then here's what will happen, in the main thread:
main will call foo.
foo will acquire the lock.
foo will call bar, which will call foo.
the 2nd foo will try to acquire the lock...