Is there any coffeescript specific trickery that would make this look neater:
index = (->
if segment == 'index'
return 0
else if segment == 'inbox'
return 2
1
)()
Yes, a switch expression:
index = switch segment
when 'index' then 0
when 'inbox' then 2
else 1
You could use an inline if ... then ... else statement broken up into multiple lines (for readability) by using the \ character.
index = if segment == 'index' then 0 \
else if segment == 'inbox' then 2 \
else 1
This is useful if your conditional logic is too complex for a simple switch block.
Yes, the CoffeeScript-specific existential operator:
index = {'index': 0, 'inbox': 2}[segment] ? 1
You could also use an inline if statement to get rid of the function call:
index = if segment == 'index' then 0 else if segment == 'inbox' then 2 else 1
But an inline if wouldn't be any harder in straight Javascript:
index = segment == 'index' ? 0 : segment == 'inbox' ? 2 : 1
Related
I am trying to use MATLAB FUNCTION block in simulink .
The model is shown below
In the "time_calc" Function i want to manipulate the variable "Sector" as shown in the code below
if sector == 1 || 2
sec = 1
elseif sector == 3 || 4
sec = 2
elseif sector == 5||6
sec = 3
elseif sector == 7||8
sec = 4
elseif sector == 9||10
sec = 5
elseif sector == 11 || 12
sec = 6
end
The below is the scope and you can see the values of "sector" changing from 0 through 12 and then repeating itself
But I am getting the value of "sec" as constant "1"(shown below in figure)
(Maybe because it is evaluating the first "1" as boolean true and running that statement only over and over again)
How to correct it ?
if sector == 1 || 2 evaluates sector == 1, if it's true, the statement is true. If it's false, it evaluates 2, which is always true, and so the statement is always true.
What you intended to write was if sector == 1 || sector == 2. You can also write this as if any(sector == [1, 2]).
Your function is equivalent to:
sec=ceil(sector/2)
#Cris Luengo's answer shows why your code is wrong. But I suggest you change the entire thing by this one liner, that is much clearer.
Remove the elseif's and replace them all with just if's
I need such a condition in ts:
([treeLevel = 0] && [globalVar = GP:R > 0]) || [PIDinRootline = {$pages.2018}]
I wanna show block if page has treelevel=0 and the get var R > 0, or if page id = $pages.2018
It looks like the similar code in php:
if(($treeLevel == 0 && $r > 0) || (pid == number))
The all expression in first brackets should be right, or in second.
Is it exist the method to group it like the previous record or I can only use userfunc?
There is no grouping in TS conditions, but if you need this particular condition from your post I think it is not needed because brackets around && are useless in this case.
(p && q) || r
is exactly the same as
p && q || r
Did you tested it?
if ((status & 0x3F) == 1 ){ }..
the status is variable in swift language.
what is mean about this condition, & mean and (status & 0x3F) value return
& is the bitwise AND operator. It compares the bits of the two operands and sets the corresponding bit to 1 if it is 1 in both operands, or to 0 if either or both are 0.
So this statement:
((status & 0x3F) == 1)
is combining status with 0b111111 (the binary equivalent of 0x3F and checking if the result is exactly 1. This will only be true if the last 6 bits of status are 0b000001.
In this if:
if( (dtc24_state[2] & 0x8) == 0x8 ) {
self.haldexABCDTC24State.text = status_str + " - UNKNOWN"
self.haldexABCDTC24State.textColor = text_color
active_or_stored_dtc = true
}
dct24_state is an array of values. The value of dct24_state[2] is combined with 0x8 or 0b1000 and checked against 0x8. This is checking if the 4th bit from the right is set. Nothing else matters. If the 4th bit from the right is set, the if is true and the code block is executed.
0x3F is 111111. So, it means this:
for each bit of yourNumber in binary system presentation use and method.
This way truncates the left part of the number. and the result compares with 1.
e.g.
7777 is 1111001100001 after executing and this number converts into
100001. So the result is false.
But for 7745 (1111001000001) the result is 1. The result is true.
The rule for 'and' function: 0 & 0 = 0 ; 0 & 1 = 0; 1 & 0 = 1; 1 & 1 = 1.
I am trying to understand Boolean logic and operators.
I found this example but can't understand why this expression will evaluate to the one shown below.
Say, a = 0, b = 1, c = 0
Expression Will Evaluate to
val1 = !(a || b || c); !(0 || 1 || 0) = !(1) = 0
As I see it, val1 is not a or not b or not c, so why it evaluates to not 1 ?
Not(a or b or c) evaluates the or operations first, so it's not the same as (not a) or (not b) or (not c).
Indeed, it's the same as (not a) AND (not b) AND (not c).
Either operand to an OR being true will give a true result, and then the NOT flips that to a false result for the expression as a whole.
As with integer or real number arithmetic, order of operation can greatly alter the result.
.... val1 is not a or not b or not c ...
No, this is incorrect. The 0 || 1 || 0 inside the parenthesis is evaluated first. The example has it right.
Let's say val1 = 1
1 = !(0 || 1 || 0)
1 = !(1) - because it is the only value that is equal to val1
1 = 0 - then it negates it afterwards
Let's go step-by-step.
val1 = !(0 || 1 || 0);
Firstly, 0 || 1 will evaluate to 1, because || means 'true if at least one of them is true, otherwise false', and 1 = true, 0 = false.
So now it is
val1 = !(1 || 0); Here 1 || 0 will again evaluate to 1, because at least one of them is 1. Now we've got val1 = !(1);. ! means the opposite of the input, so !(1) = 0.
As I see it, val1 is not a or not b or not c, so why it evaluates to not 1 ?
Because what you say would be written as val1 = !0 || !1 || !0. Its quite different, because it doesn't have parenthesis. Parenthesis means 'evaluate everything in the parenthesis first'.
In the code below, from a blog post by Alias, I noticed the use of the double exclamation mark !!. I was wondering what it meant and where I could go in the future to find explanations for Perl syntax like this. (Yes, I already searched for !! at perlsyn).
package Foo;
use vars qw{$DEBUG};
BEGIN {
$DEBUG = 0 unless defined $DEBUG;
}
use constant DEBUG => !! $DEBUG;
sub foo {
debug('In sub foo') if DEBUG;
...
}
UPDATE
Thanks for all of your answers.
Here is something else I just found that is related The List Squash Operator x!!
It is just two ! boolean not operators sitting next to each other.
The reason to use this idiom is to make sure that you receive a 1 or a 0. Actually it returns an empty string which numifys to 0. It's usually only used in numeric, or boolean context though.
You will often see this in Code Golf competitions, because it is shorter than using the ternary ? : operator with 1 and 0 ($test ? 1 : 0).
!! undef == 0
!! 0 == 0
!! 1 == 1
!! $obj == 1
!! 100 == 1
undef ? 1 : 0 == 0
0 ? 1 : 0 == 0
1 ? 1 : 0 == 1
$obj ? 1 : 0 == 1
100 ? 1 : 0 == 1
not-not.
It converts the value to a boolean (or as close as Perl gets to such a thing).
Because three other answers claim that the range is "0" or "1", I just thought I'd mention that booleans in Perl (as returned by operators like ==, not, and so on) are undef and 1, not 0 and 1.