Not sure what the value of a is - php

where:
$b = true;
$c = 0;
$a = ($a ? ($a ? $b : $c) : ($c ? $a : $b));
I'm not sure how to work out a.
So I understand that this is a shorthand operator, and usually it's a case of:
$value ? true : false
meaning
if $a = true { true } else { false };
so:
if $a{
if $a{
true;}
else{
0;};
else{
if $0{
$a;}
else{
true;}
};
does this make the value of $a true?

The value of $a would be true
$b = true;
$c = 0;
$a = ($a ? ($a ? $b : $c) : ($c ? $a : $b));
The shorthand can be interpreted like this:
if($a) {
if($a) {
$a = $b;
} else {
$a = $c;
}
} else {
if($c) {
$a = $a;
} else {
$a = $b;
}
}
Because $a is false for not existing in the first place, it immediately jumps to the else statement in that. So the only part that matters to you is:
if($c) {
$a = $a;
} else {
$a = $b;
}
0 is the same as false, so $c will come back as false, therefore $a is equal to $b, which is true.
Edit:
There is some discussion on the notice that is thrown, but this fails to account for the fact that notices are not truly errors and because of this there is no interruption to the code. The result is not Notice: Undefined variable: a, the "result" (think these people mean output) would be blank if it weren't for us determining the value of $a at the end with var_dump. The question was as to what the value of $a becomes, not what appears on your screen.
Something displaying on your screen in re to a variable not being set has nothing to do with the value of what $a is.
If you execute the following code, the notice is not the only thing realized:
$b = true;
$c = 0;
$a = ($a ? ($a ? $b : $c) : ($c ? $a : $b));
var_dump($a);
So the output is:
E_NOTICE : type 8 -- Undefined variable: a -- at line 5
bool(true)
The fact that a notice was thrown does not prevent $a from becoming true.
Also notices are easily suppressed...
error_reporting(0);
$b = true;
$c = 0;
$a = ($a ? ($a ? $b : $c) : ($c ? $a : $b));
var_dump($a);
would result in $a still becoming true, and without seeing the notice.
bool(true)

If you run the code as is, you would get: Notice: Undefined variable: a in myfile.php on line 4
Therefore, I would postulate $a is set somewhere earlier. Yet, whatever value $a has prior, if $a is can be evaluated to true or false, $a would still be true after running your code for the following reason:
If $a were true, then the first part would yield $a = $b and we know $b = true.
if(TRUE) {
if(TRUE) {
$a = $b; //AND $b == TRUE
} else {
$a = $c;
}
} else {
...
}
If $a were false, then the second part would yield $a = $b again
if(FALSE) {
...
} else {
if(0) { // 0 will equate to FALSE
...
} else {
// 0 is the same as FALSE so we end up again with $a = $b
$a = $b; //AND $b == TRUE
}
}
In fact, if you run this code, it will show you the value of $a is true both times:
<?php
$a = false;
$b = true;
$c = 0;
$a = ($a ? ($a ? $b : $c) : ($c ? $a : $b));
echo $a;
$a = true;
$b = true;
$c = 0;
$a = ($a ? ($a ? $b : $c) : ($c ? $a : $b));
echo $a;

Related

OR Preference changing with a return

From what I see operator precedence makes sense in these two examples:
$a = false;
$b = true;
$c = $a || $b;
Here $c is true
$a = false;
$b = true;
$c = $a or $b;
Here $c is false
I understand the reasoning behind it. However the following:
$a = false;
$b = true;
return $a or $b;
Returns true, which puzzles me.
What is the reason for this?
or has lower precedence than =, so this:
$c = $a or $b;
Becomes this:
($c = $a) or $b;
But this doesn't make sense:
(return $a) or $b;
So you get this:
return ($a or $b);
Within an expression, operator precedence applies. =, || and or are all operators and $c = $a or $b is an expression. And according to operator precedence it evaluates as ($c = $a) or $b.
However, return is a statement. return is not an operator and does not group by operator precedence. It always evaluates as return <expression>, and therefore always as return ($a or $b).
The result of the expression $c = $a or $b is true BTW. $c is being assigned false in the course of the expression, but the expression overall returns the value true ($b). So even this would return true:
return $c = $a or $b;

PHP use result of as variable

Is it possible to use the result of an if with an OR statement as a variable for a function?
As example:
$a = true;
$b = false;
if ($a || $b) {
$this->functionCall($a)
}
Other example:
$a = false;
$b = true;
if ($a || $b) {
$this->functionCall($b)
}
Third and final exmaple:
$a = true;
$b = true;
if ($a || $b) {
$this->functionCall($a, $b)
}
So I need to detect what variable is true and pass it as a paramater. Is this even possible?
Any helps is appreciated!
Many thanks in advance
I'd do the logic bit inside a two-parameter function if I were you, as such :
function myFunc($a = false, $b = false) {
if ($a == true)
echo 'a';
if ($b == true)
echo 'b';
}
myFunc(); // echoes nothing
$a = true;
$b = false;
myFunc($a, $b); // echoes 'a'
$a = false;
$b = true;
myFunc($a, $b); // echoes 'b'
$a = true;
$b = true;
myFunc($a, $b); // echoes 'ab'
PHP 5.6+ version, filter out the falsely values (you can pass a callback to array_filter for different checks) and use those with the splat operator.
$params = array_filter([$a, $b]);
$this->callFunction(...$params);
No need for any IF checks and confusing in IF assignments.
Explore Variadic functions and Argument unpacking.

Do you know a nice PHP hack to do inline conditional assignation?

I really enjoy || operator in JavaScript, where we can do inline conditional assignation.
var a = 0;
var b = 42;
var test = a || b || 'default value';
console.log(test); // 42
This is clear to read, and don't take too many lines.
In PHP, this logical operator return booleans:
$a = 0;
$b = 42;
$test = $a || $b || 'default value';
print_r($test); // bool(true)
Of course, we can do inline assignation using ternaries:
$test = $a ? $a : $b ? $b : 'default';
print_r($test); // int(42)
But this make code ambiguous, this is not that easy to read.
So here my question come:
Do you know a nice PHP hack to do inline conditional assignation ?
In PHP 5.3+ you can do this:
$test = $a ?: ($b ?: 'default value');
This will work as long as you don't need to short-circuit side effects:
function either_or() {
$nargs = func_num_args();
if ($nargs == 0) {
return false;
}
$args = func_get_args();
for ($i = 0; $i < $nargs-1; $i++) {
if ($args[$i]) {
return $args[$i];
}
}
return $args[$nargs-1];
}
$test = either_or($a, $b, "Default value");

How to find the highest combination while comparing three numbers in php

There are three numbers in set A {3,4,7} and in set B {2,4,7}.
It is not possible to get the result true because the the first numbers in a and b are not same.
But I need to get the result as true by comparing other two number and leaving the first number.
How is it possible to do it in PHP?
try this it will helps you.. it will display the 1 => for more number of combinations as same and 0 => for more of combination as different . in the below code the more number of combination as different so its returns 0.
<?php
$A = array(2,3,7,5,6);
$B = array(4,3,7,8,9);
$flagTrue = 0;
$flagFalse = 0;
for($i=0; $i < count($A); $i++)
{
if($A[$i] == $B[$i])
{
$flagTrue=$flagTrue+1;
}
else
{
$flagFalse=$flagFalse+1;
}
}
$var_is_greater_than_two = ($flagTrue >= $flagFalse) ? 1 : 0;
echo $var_is_greater_than_two;
?>
<?php
$a = array(3,4,7);
$b = array(2,4,7);
echo $a === $b ? 'TRUE' : 'FALSE';
echo PHP_EOL;
array_shift($a);
array_shift($b);
echo $a === $b ? 'TRUE' : 'FALSE';
?>
Shows:
FALSE TRUE
UPD:
If you need to extract values from strings, then:
$strA = '3,4,7';
$strB = '2,4,7';
$a = explode(',', $strA);
$b = explode(',', $strB);
array_shift($a);
array_shift($b);
echo $a === $b ? 'TRUE' : 'FALSE';
Should work.
<?
$A = array(2,3,7);
$B = array(4,3,7);
$isTrue=1;
for($i=1; $i < count($A); $i++) if($A[$i]!=$B[$i]) $isTrue=0;
echo $isTrue;
?>
EDIT:
If you want to return true if exactly two elements are the same, then the code would be:
$common=0;
for($i=0; $i < count($a); $i++) if($a[$i]==$b[$i]) $common++;
if($common==2) $isTrue=1;
function compareSets($a, $b) {
$result = TRUE;
$diffArray = array_diff($a, $b);
foreach ($diffArray as $key => $value) {
if ($key > 0) {
$result = FALSE;
break;
}
}
return $result;
}
<?php
$a = array(3,4,7);
$b = array(2,4,7);
for($i=0;$i<3;$i++)
{
if($a[$i] > $b[$i]
echo true;
}
?>
will give true if a is greater.

Accessing a variable from a function inside a function?

I have the following code:
function create_sort_callback($criteria)
{
return function($a, $b)
{
$a = $a[$criteria];
$b = $b[$criteria];
return ($a == $b ? 0 : (($a < $b) ? -1 : 1));
};
}
It turns out I can't access $criteria from within the inner function. How can I solve this problem?
Try like this
function create_sort_callback($criteria)
{
return function($a, $b) use($criteria)
{
$a = $a[$criteria];
$b = $b[$criteria];
return ($a == $b ? 0 : (($a < $b) ? -1 : 1));
};
}
You need using closures http://www.php.net/manual/en/functions.anonymous.php
Use the use keyword.
function create_sort_callback($criteria)
{
return function($a, $b) use ($criteria)
{
$a = $a[$criteria];
$b = $b[$criteria];
return ($a == $b ? 0 : (($a < $b) ? -1 : 1));
};
}

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