PHP:
$a = 2;
$b = 3;
if($b=1 && $a=5)
{
$a++;
$b++;
}
echo $a.'-'.$b;
$a = 2;
$b = 3;
if($a=5 and $b=1)
{
$a++;
$b++;
}
echo $a.'-'.$b;
Output 6-16-2.I don't understand the 1 here.
Perl :
$a = 2;
$b = 3;
if($b=1 && $a=5)
{
$a++;
$b++;
}
print $a.'-'.$b;
$a = 2;
$b = 3;
if($a=5 and $b=1)
{
$a++;
$b++;
}
print $a.'-'.$b;
Output 6-66-2, I don't understand the second 6 here.
Anyone knows the reason?
Actually I know && has higher precedence than and,but I still has the doubt when knowing this before hand.
UPDATE
Now I understand the PHP one,what about the Perl one?
Regarding Perl:
Unlike PHP (but like Python, JavaScript, etc.) the boolean operators don't return a boolean value but the value that made the expression true (or the last value) determines the final result of the expression† (source).
$b=1 && $a=5
is evaluated as
$b = (1 && $a=5) // same as in PHP
which is the same as $b = (1 && 5) (assignment "returns" the assigned value) and assigns 5 to $b.
The bottom line is: The operator precedence is the same in Perl and PHP (at least in this case), but they differ in what value is returned by the boolean operators.
FWIW, PHP's operator precedence can be found here.
What's more interesting (at least this was new to me) is that PHP does not perform type conversion for the increment/decrement operators.
So if $b is true, then $b++ leaves the value as true, while e.g. $b += 1 assigns 2 to $b.
†: What I mean with this is that it returns the first (leftmost) value which
evaluates to false in case of &&
evaluates to true in case of ||
or the last value of the expression.
First example
$a = 2;
$b = 3;
if($b=1 && $a=5) // means $b = (1 && $a=5)
{
var_dump($b); //bool(true) because of &&
$a++;
$b++; //bool(true)++ ==true, ok
}
echo $a.'-'.$b;
hope you will not use those codes in production)
I'm noob in perl but i can suggest a&&b returns a or b (last of them if all of them converted to bool), not boolean, then $b = (1 && $a=5) returns $b=5 (is 5)
here's the issue: 1 && 5 returns 5 in perl. you get the result you expect if you code the conditional as if(($b=1) && ($a=5))
For Perl, fig. 2: and has a very low priority in perl, it's not a synonym of &&'s. Therefore the sample is executed as (($a = 5) and ($b = 1)) which sets $a and $b to 5 and 1 respectively and returns a value of the last argument (i.e. 1).
After ++'s you get 6-2.
refer to http://sillythingsthatmatter.in/PHP/operators.php for good examples
Related
This question already has answers here:
What's the difference between ++$i and $i++ in PHP?
(15 answers)
Closed 7 years ago.
$a = 3;
$b = $a++;
if ($a > $b) { echo “a > $b” }
else if ($a == $b) { echo “a = $b” }
else { echo “a < $b” }
When I work through this question, I get a=3, b=4 (3+1). Hence both the If and Else If conditions are false, so I go to Else and final answer is: a < 4.
However, the answer according to the mark scheme is: a > 3 meaning that the If condition is true. How could $a be larger than $b? Thanks
Take a look at the following statement:
$b = $a++;
The ++ is located after $a. This is the post-increment operator.
It first returns the current value of $a (3), and only then increments $a. In other words, $b is assigned with $a's current value, and then $a is incremented. So, $a is 4 and $b is 3, hence $a > $b.
If I have a conditional statement
if (A > B || B > C)
Which statement is going to be evaluated first: "A > B" or "B > C"?
Does same order is applied to math statements:
$var = $value1 + $value2 + $value3;
Thanks,
Alex.
In PHP the script is evaluated from left to right unless parenthesis are used, if they are used it evalutes then in logical order. In addition please remember that no code in the if condition block(including evaluators) are ran past the first failing statement. This example will only execute the second echo $a and it's value will be 0
$a = 0;
if(1 == 0 && $a = 5)
{
echo $a;
}
echo $a;
This statement will have $a value of 5 and will execute the statement. Interestingly, the reason that the code will execute is because the $a = 5 assignment in the if sets $a = 5 or " 5 = 5".
if(1 == 1 && $a = 5)
{
echo $a;
}
Also note there are else and else if statements if you have not looked into it
$a =2
if($a == 2)
{
}
else if($a > 2){
echo ">".$a;
}
else{
echo "its none of the conditions";
}
The reason that you use two equals signs is to compare the value type insensitive vs one equal which would be assigning the value. There is also three equals which would compare the type and value example This would evaluate to true :
$a = 2;
if($a == "2")
The following would not be true because you are comparing a String to integer.
$a = 2;
if($a === "2")
Regarding your second questions the same is true of String operators but your syntax is INVALID.
This Example Would say Hellow World:
echo "hellow"."world";
This Example IS NOT DOING CONCATENATION(Though it would do addition if they are integers)
echo "hellow" + "world";
I am using logical operators to test variables but AND & operator work fine but OR | and Either-OR ^ always true.
Why?
$a = 6;
$b = 6;
if ($a OR $b == 3) {
echo 'true <br />';
}
else {
echo 'false <br />';
}
The issue is with your syntax.
You need to look at the expression separately.
if($a) OR if($b == 3)
is what you're doing.
What you want is:
if($a == 3 || $b == 3)
If you look at $a by itself, any value except for 0 will return true making the entire equation true thanks to the OR
Because you have to OR to Boolean results - you're reading it too much like English.
if ($a == 3 || $b == 3)
rather than
if ($a OR $b == 3)
It's a matter of precedence - see http://php.net/manual/en/language.operators.precedence.php for more details here.
Both the other answers give you the code you need.
$a = true
$b = true
if($a and $b) TRUE if both $a and $b are TRUE.
Reference: http://php.net/manual/en/language.operators.logical.php
It is my first question, so hi all..
Is it possible to satisfy this condition?
Where $a and $b are some defined variables
if(!is_numeric($a * $b.'')) { ... }
EDIT: Second conditon
$a * $b can't be INF
You can satisfy the condition with specific values of $a and $b. For example, $a = $b = 1e200, or $a = 1e400, $b = 0.
If the product of $a and $b overflows to infinity or is a not-a-number, the concatenation with an empty string will produce the string INF or NAN:
php > var_export(1e200*1e200);
INF
php > var_export(is_numeric(1e200*1e200));
true
php > var_export(is_numeric(1e200*1e200.''));
false
php > var_export(1e400*0);
NAN
php > var_export(is_numeric(1e400*0));
true
php > var_export(is_numeric(1e400*0 .''));
false
Tested using 32-bit PHP
$a = 9.9E300;
$b = 9.9E300;
if(!is_numeric($a * $b.'')) {
echo 'non-numeric';
} else {
echo 'numeric';
}
I'm studying for my finals and I came across this question:
consider this following PHP code, write the output after executing it
<?php
$a=3;
$b=$a++;
IF($a>$b)
{
echo "a>$b";
}
else if ($a == $b)
{
echo "a=$b";
}
else
{
echo "a < $b";
}
?>
When I output it in my text editor I get a < 3, but I don't understand why though?
I thought a is assigned to 3 and also b is assigned to a++ 3 and 3==3 so should a==3 be printed out?
No, you are using post-increment operator on $a. So, $b will be assigned a value of 3, and later, when the statement is executed, $a will increment itself by one, and become 4. So, you'll now be comparing $a as 4 and $b as 3.
Hence you get the result a > 3
The $a++ incrementation happens after the expression gets evaluated, while ++$a would happen before.
So in your case, $b was first set to 3, and then $a was increased.
$a++ tells the variable $a explicitely to increase, no matter if you assigning to another variable or not!
This gives the possibility to do things like if ($a++ > 10) { // ... in loops.
For your case you would have to take $b = $a + 1;
<?php
$a=3;
$b=$a++;
// $b = 3 and $a = 4 now
IF($a>$b)
{
echo "a>$b";
}
else if ($a == $b)
{
echo "a=$b";
}
else
{
echo "a < $b";
}
?>
I tested your code and I get:
a>3
which makes sense
$a is 3 but is increased to 4 when you do $a++
$b is just $a before the ++ action so it stays 3
Think of $a++ as $a = $a + 1 then it makes sense