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Does a break statement break from a switch/select?
...ully illustrative example:
loop:
for {
switch expr {
case foo:
if condA {
doA()
break // like 'goto A'
}
if condB {
doB()
break loop // li...
How to click first link in list of items after upgrading to Capybara 2.0?
...ge.find("div.panel", text: /Proposals/) do
within page.find('tr', text: /Foo/) do
page.should have_xpath('td[3]', text: @today)
end
end
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...
Aligning UIToolBar items
...e: "A", style: UIBarButtonItemStyle.Plain, target: viewController, action: foo)
let button2 = UIBarButtonItem(title: "B", style: UIBarButtonItemStyle.Plain, target: viewController, action: bar)
let button3 = UIBarButtonItem(title: "C", style: UIBarButtonItemStyle.Plain, target: viewControlle...
JavaScript replace/regex
...he regexp):
"$TESTONE $TESTONE".replace( new RegExp("\\$TESTONE","gm"),"foo")
Otherwise, it looks for the end of the line and 'TESTONE' (which it never finds).
Personally, I'm not a big fan of building regexp's using strings for this reason. The level of escaping that's needed could lead you ...
Difference between `mod` and `rem` in Haskell
...nswered May 4 '11 at 23:54
Fred FooFred Foo
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Do you need break in switch when return is used?
...ion execution.
Also, if all of your case statements are like this:
case 'foo':
$result = find_result(...);
break;
And after the switch statement you just have return $result, using return find_result(...); in each case will make your code much more readable.
Lastly, don't forget to add th...
How does one use rescue in Ruby without the begin and end block
...
A method "def" can serve as a "begin" statement:
def foo
...
rescue
...
end
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How do you get the index of the current iteration of a foreach loop?
...e
{
public static void Main()
{
string[] values = new[] { "foo", "bar", "baz" };
values.ForEachWithIndex((item, idx) => Console.WriteLine("{0}: {1}", idx, item));
}
}
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How do I show the value of a #define at compile-time?
...code using "BOOST_VERSION"? Perhaps something like "blah[BOOST_VERSION] = foo;" will tell you something like "string literal 1.2.1 cannot be used as an array address". It won't be a pretty error message, but at least it'll show you the relevant value. You can play around until you find a compile ...
How to apply a function to two columns of Pandas dataframe
...like sending 2 columns and a constant into a function (i.e. col_1, col_2, 'foo').
import numpy as np
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
#df['col_3...
