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Nullable type issue with ?: Conditional Operator
Could someone explain why this works in C#.NET 2.0:
5 Answers
5
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How did Google manage to do this? Slide ActionBar in Android application
...
150
In fact, there's a way to do this. Even without implementing your own ActionBar.
Just have a lo...
Check if an element's content is overflowing?
... Your css looks like this:
.scrollbox {
overflow: auto;
width: 200px;
max-height: 200px;
margin: 50px auto;
background:
/* Shadow covers */
linear-gradient(white 30%, rgba(255,255,255,0)),
linear-gradient(rgba(255,255,255,0), white 70%) 0 100%,
/* Shad...
Full Page
...ks for everyone responding.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Test Layout</title>
<style type="text/...
Is there an expression for an infinite generator?
...gument iter = zero-argument callable + sentinel value
int() always returns 0
Therefore, iter(int, 1) is an infinite iterator. There are obviously a huge number of variations on this particular theme (especially once you add lambda into the mix). One variant of particular note is iter(f, object()),...
Mercurial move changes to a new branch
...e extension. Suppose you have a history like this:
@ changeset: 2:81b92083cb1d
| tag: tip
| summary: my new feature: edit file a
|
o changeset: 1:8bdc4508ac7b
| summary: my new feature: add file b
|
o changeset: 0:d554afd54164
summary: initial
This means, revisi...
Difference between Big-O and Little-O Notation
... O(g) says, essentially
For at least one choice of a constant k > 0, you can find a constant a such that the inequality 0 <= f(x) <= k g(x) holds for all x > a.
Note that O(g) is the set of all functions for which this condition holds.
f ∈ o(g) says, essentially
For every c...
Why do enum permissions often have 0, 1, 2, 4 values?
Why are people always using enum values like 0, 1, 2, 4, 8 and not 0, 1, 2, 3, 4 ?
7 Answers
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How does variable assignment work in JavaScript?
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edited Aug 11 '18 at 14:50
Volker E.
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answered...