As far as I know the "tag" field in UIView is integer.
Why one of my UIButtons has "nil" value in it's tag?
(lldb) po [button_one tag]
nil
I've set the tag 0 in designer, but still returns nil tag in code.
po means 'print object'. An integer isn't an object. And nil would be equal to an integer value of zero.
So, in lldb you should be using p to print the integer value. And it is probably correct.
My output is slightly different, but the point is, trying to print object on an integer (as Wain says) results in nil if the integer is 0 and an error if it is non-zero. (but in this case lldb still gives the p (int) value)
(lldb) po 0
$0 = 0 <nil>
(lldb) po nil
$1 = 0x00000000 <nil>
(lldb) po button.tag
$2 = 0 <nil>
(lldb) p (int)button.tag
(int) $3 = 0
Set the tag to 1
(lldb) po button.tag
$4 = 1 [no Objective-C description available]
(lldb) p (int)button.tag
(int) $5 = 1
Related
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'.
Does anyone know if there is a way to use some kind shorthand in swift? more specifically, leaving out the braces in things like IF statements... eg
if num == 0
// Do something
instead of
if num == 0
{
// Do something
}
Those braces become rather space consuming when you have a few nested IF's.
PS. I do know I can do the following:
if num == 0 {
// Do something }
But I'm still curious if that sort of thing is possible
You can do that :
let x = 10, y = 20;
let max = (x < y) ? y : x ; // So max = 20
And so much interesting things :
let max = (x < y) ? "y is greater than x" : "x is greater than y" // max = "y is greater than x"
let max = (x < y) ? true : false // max = true
let max = (x > y) ? func() : anotherFunc() // max = anotherFunc()
(x < y) ? func() : anotherFunc() // code is running func()
This following stack : http://codereview.stackexchange.com can be better for your question ;)
Edit : ternary operators and compilation
By doing nothing more than replacing the ternary operator with an if else statement, the build time was reduced by 92.9%.
https://medium.com/#RobertGummesson/regarding-swift-build-time-optimizations-fc92cdd91e31#.42uncapwc
In swift you have to add braces even if there is just one statement in if:
if num == 0 {
// Do something
}
You cannot leave the braces, that how swift if statement work.
You could use a shorthand if statement like you would in objective-c:
num1 < num2 ? DO SOMETHING IF TRUE : DO SOMETHING IF FALSE
Swift 2.0 update
Method 1:
a != nil ? a! : b
Method 2: Shorthand if
b = a ?? ""
Referance: Apple Docs: Ternary Conditional Operator
and it does work,
u.dob = (userInfo["dob"] as? String) != nil ? (userInfo["dob"] as! String):""
I am replacing a json string with blank string if it is nil.
Edit: Adding Gerardo Medina`s suggestion...we can always use shorthand If
u.dob = userInfo["dob"] as? String ?? ""
It is called shorthand if-else condition. If you are into iOS development in Swift, then you can also manipulate your UI objects' behaviour with this property.
For e.g. - I want my button to be enabled only when there is some text in the textfield. In other words, should stay disabled when character count in textfield is zero.
button.enabled = (textField.characters.count > 0) ? true : false
its very simple :
in Swift 4
playButton.currentTitle == "Play" ? startPlay() : stopPlay()
Original Code is
if playButton.currentTitle == "Play"{
StartPlay()
}else{
StopPlay()
}
You could always put the entire if on one line:
if num == 0 { temp = 0 }
I have the following file
...
MODE P E
IMP:P 1 19r 0
IMP:E 1 19r 0
...
SDEF POS= 0 0 14.6 AXS= 0 0 1 EXT=d3 RAD= d4 cell=23 ERG=d1 PAR=2
SI1 L 0.020
SP1 1
SI4 0. 3.401
SI3 0.9
...
NPS 20000000
I want to do the following task
Check if after the sequence ERG= there is a number or a string.
If it's a string, find the sequence SI1 L and change the value after that, using values that the user inputs.
If it's a number, change the number using values that the user inputs.
Note that if after ERG= there is a number, there will be no SI1 L sequence.
For instance number 2 can be accomplished using the following
#! /bin/bash
vals=(0.02 0.03 0.04 0.05)
for val in "${vals[#]}"; do
awk -vval="$val" '$1=="SI1"{$3=val}1' 20
done
How can the above algorithm be achieved?
#!/bin/bash
val="$#"
awk -v val="$val" '
BEGIN { i=1; split (val,v," ") }
# If it is a string, find the sequence SI1 L and change the value after that, using values that the user inputs
/SDEF POS.*ERG=[a-zA-Z]+/ { flag="y" ; }
/SI1 L/ { if (flag=="y") { $3=v[i]; i++; flag="n"; } }
# If it is a number, change the number using values that the user inputs.
/SDEF POS.*ERG=[0-9]+ / { sub(/ERG=[0-9]*/, "ERG="v[i],$0);i++; }
1
' file
hints:
If the rule find ERG with at least one or more letters ([a-zA-Z]+, it will set the flag.
The /SI1 L/ rule will only triggers, if the flag is set. If the rule triggered, it would unset the flag again, so that any following /SI L/ wouldn't trigger again.
.* stands for 0-n sign or character
[A-Za-z]+ stands for 1-n alphabetic character in lower or upper case
awk -F '[[:blank:]=]' -v string_value="foo" -v number_value=42 '
/ERG=/ {
for (i=1; i<NF; i++)
if ($i == "ERG") {
isstring = ($(i+1) ~ /[^[:digit:]]/)
break
}
if (!isstring)
$(i+1) = number_value
}
/SI1 L/ && isstring { $NF = string_value }
1
' filename
In the code below, from a blog post by Alias, I noticed the use of the double exclamation mark !!. I was wondering what it meant and where I could go in the future to find explanations for Perl syntax like this. (Yes, I already searched for !! at perlsyn).
package Foo;
use vars qw{$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.