Consider this code:
$x = 1.4;
$i1 = 0.5;
$i2 = 0.4;
echo ($i1 + $i2 = $x); // Outputs 1.9
Why is this? I've tried searching for this kind of variable setup without results. Is the variable $i2 being ignored? Why use this over echo ($x + $i1);? It outputs the same result.
The point is that it does two things in one statement.
It is shorthand for:
$i2 = $x;
echo ($i1 + $i2);
The assignment happens inline, saving the separate line. Not ideal style, but often used in if(), while() and other control statements.
that would be $i1 + the assignment.
The assignment evaluates to $x ($i2 = $x )
the end result is echo 0.5 + 1.4.
Even php has operator priorities http://php.net/manual/en/language.operators.precedence.php.
= is treated before +, which means that this is what happens:
echo ($i1 + ($i2 = $x));
Related
Take this example:
<?php
$i = 1;
$i += 2;
$i =+ 5;
echo $i; // 5
This shows that =+ is an assignment operator. Still, this is highly confusing to me and not at all semantic. I spent hours debugging something simply because I accidentally used =+ instead of +=. The former does not throw errors, though. So I am curious: what is the use case for =+. When would you ever (need to) use it over a simple =?
=+ is not an operator. Think of it as
$i = +5;
as opposed to
$i = -5;
=+ is not a single operator. Its two: the assignment (=) and the unary plus (+).
If you insert different whitespace it gets obvious:
$i = +5;
The unary plus in PHP is used for conversion of $i to int or float as appropriate. Since 5 already is an int, its the identity in this case, and the whole expression is semantically equivalent to:
$i = 5;
As you mentioned += is an assignment operator is correct.
In your case:
$i = 1;
Just assign the value 1 to $i and then when your second line execute $i += 2;, its just add the value 2 with the existing value of $i.
So now when you use $i =+ 5; in your third line, its re-assign the variable $i with the positive value +5.
There are no big deal with =+ and += these are different.
How to solve nth degree equations in PHP
Example:
1/(1+i)+1/(1+i)2+...1/(1+i)n=k
While k is the constant,I'd like to find value of i.
How can I achieve this in PHP?
First of all, your expression on the left is a geometric sum, so you can rewrite it as (using x=1+i)
1/x*(1+...+1/x^(n-1)) = 1/x * (1-1/x^n)/(1-1/x) = (1-x^(-n))/(x-1)
and consequently the equation can be rewritten as
(1 - pow( 1+i, -n))/i = k
Now from the original expression one knows that the left side as a sum of convex monotonically decreasing functions is equally so, thus any of bisection, regula falsi variants or secant method will work sufficiently well.
Use
(1+i)^(-n)=1 - n*i + (n*(n+1))/2*i^2 +...
to get the approximative equation and first approximation
1-(n+1)/2*i = k/n <=> i = (1-k/n)*2/(n+1)
so that you can start bracketing method with the interval from 0 to twice this i.
Try something like this....
$n = 5;
$i = 2;
$k = null;
for ($x = 1; $x <= $n; $x++) {
$k += 1 / pow((1 + $i), $x);
}
echo $k; //Answer --> 0.49794238683128
Why doesn't the double equals work in a PHP for loop? The following works:
$cnt = 5;
for ($z = 0; $z <= $cnt; $z++) {
echo $z."!<br />";
}
But why doesn't this work?
$cnt = 5;
for ($z = 0; $z == $cnt; $z++) {
echo $z."!<br />";
}
The loop executes only if the condition evaluates to true. At the first iteration, $z == $cnt is false, so the loop never executes.
A common loop strategy is to use a sentinel value:
$cnt = 5;
$stop = $cnt + 1;
for ($z = 0; $z != $stop; $z++) {
. . .
}
Note that the comparison is negative (!= or !==). Using a sentinel is usually unnecessary for numerical loop variables (I wouldn't recommend it for your posted code), but is useful for other situations (e.g., when $stop represents null, an illegal value, etc.). It's particularly helpful when the loop variable changes in a pattern that is not easy to characterize succinctly.
because on the first iteration $z != $cnt, so the loop stops immediately.
Let us look at this from the computer's perspective:
If I am a computer this is what you told me to do:
set $cnt = 5;
set $z = 0;
check if $z equals $cnt
if so do whatever is in the loop, then increment $z
Trouble is, 5 does not equal 0 and never will, so the loop will simply be skipped. If you had $cnt = $z+1 inside the loop this would be an infinite loop.
So, you see, == works just fine, it simply doesn't do what you think it should do.
Hope this helps!
A for loop works by looping until a condition is made false. In the first example the loop will execute until z = 5 after that z will not longer be less than or equal to cnt (z starts at 0 and increments through each loop). In the second example you have $z == $cnt as your condition, and since z = 0 and cnt = 5 the loop will stop automatically because the condition is made false. You can use not equals instead like following:
$cnt = 6;
for ($z = 0; $z != $cnt; $z++) {
echo $z."!<br />";
}
The syntax of for is:
for (<init>; <condition>; <increment>)
The loop tests <condition> before each iteration. If it's true, it executes that iteration; if it's false, the loop terminates.
In your case, since $z == $cnt is false before the first iteration, the loop terminates immediately.
To do what you want, invert the test. You also need to bump the end value up by one, since the original version used <=, not <. Note that in both cases, the loop executes $cnt+1 times, because you start from 0.
for ($z = 0; $z != $cnt+1; $z++)
How are they different? Here's what I'm thinking, but I'm not sure....
If you use pre-incrementation, for example in a for loop with ++j, then you are basically saying: "Make a copy of the value of j for use in the loop, then increment j, then go through the statements in the loop with the copy of j." If you are using post-incrementation in the same loop j++, then you are basically saying: "Make a copy of the value of j for use in the loop, then go through the statements in the loop with the copy of j, then increment j."
The reason I'm unsure is because I've created a for loop that multiplies the value of j by 10 and then outputs the result for j=1 through j=12, using both post- and pre-incrementation. The human readable output is exactly the same with post- and pre-incrementation. I'm thinking, 'How are the outputs exactly the same if there isn't some kind of copy operation involved?'
So, I'm guessing the difference between pre- and post-incrementation truly becomes important, in php, when I use references (which act as pointers in php) rather than names for return values? This would be because copies of references aren't made, so pre-incrementation would be: "Increment j, then go through the statements in the loop with the changed value of j, then increment j again...," whereas post-incremetation would look like: "Use the value of j for the statements in the loop, then change the value of j, then go through the loop with the new value of j..."
Pre- or post-incrementing do not magically delay things until later. It's simply inline shorthand.
// pre-increment
$var = 5;
print(++$var); // increments first, then passes value (now 6) to print()
// post-increment
$var = 5;
print($var++); // passes value (still 5) to print(), then increments
Now let's look at a loop.
for ($i = 0; $i < 9; $i++) {
print($i);
}
The last part of the loop declaration (the $i++) is simply the statement to execute after each time through the loop. It "passes" the value to nowhere, then increments it. $i isn't used anywhere at that time. Later when the next statement is executed (print($i);), the value of $i has already increased.
// add 1, then do nothing with $i
for ($i = 0; $i < 9; ++$i) {}
// do nothing with $i, then add 1
for ($i = 0; $i < 9; $i++) {}
Whichever way you do it, $i will be the same within the loop.
If it helps, you can think of them as small routines that kind of do this:
// ++$i
{
$i = $i + 1;
return $i;
}
// $i++
{
return $i;
$i = $i + 1;
}
As I reread your question, I think the confusion is more with how the loop works than how increment operators work. Keeping in mind that the increment is a straightforward, all-at-once operation, here's how third expression in the loop works.
// here's a basic loop
for ($i = 0; $i < 9; $i++) {
// do loop stuff
print($i);
}
// this is exactly what happens
for ($i = 0; $i < 9; ) {
// do loop stuff
print($i);
$i++;
}
Just because that last line can be put in the loop declaration doesn't give it any special powers. There are no references or anything used behind the scenes. The same $i variable is seen both inside and outside the loop. Every statement inside or outside the loop directly looks up the value of $i when necessary. That's it. No funny business.
When doing $x++, you are post-incrementing... This means that the incrementation will only occur after the statement has been evaluated.
So, given the following code:
$x = 10; $y = 0; $z = 5;
$y = $z * $x++;
PHP does this:
$x = 10; $y = 0; $z = 5;
$y = $z * $x++;
// Ignore Post-Increment, Evalutate
$y = $z * $x;
$y = 5 * 10;
// Now Increment x - POST-INCREMENT
$x = $x + 1;
$x = 10 + 1;
$x = 11;
// Continue evaluating statement
$y = 5 * 10;
$y = 50;
When doing ++$x, you are pre-incrementing... This means that the incrementation will occur before the statement is evaluated:
$x = 10; $y = 0; $z = 5;
$y = $z * ++$x;
// Do Pre-Increment
$x = $x + 1;
$x = 10 + 1;
$x = 11;
// Evaluate
$y = $z * $x;
$y = 5 * 11;
$y = 55;
In the case of a for loop in PHP, PHP evaluates a for loop as follows:
for($i = 0; $i < 30; $i++) {
doSomething();
}
// Is evaluated EXACTLY as such by PHP
$i = 0;
while($i < 30) {
doSomething();
$i++;
}
The first expression ($i = 0) is evaluated (executed) once unconditionally at the beginning of the loop.
In the beginning of each iteration, $i < 30 is evaluated. If it evaluates to TRUE, the loop continues and the nested statement(s) are executed. If it evaluates to FALSE, the execution of the loop ends.
At the end of each iteration, $i++ is evaluated (executed) as an independent expression.
Therefore, post-incrementing or pre-incrementing a variable as the third expression in the loop doesn't have an effect on the behavior of it. In this simple case, both expressions will behave exactly the same.
However, in a complex loop such as the following:
for($i = $j = 0; $i < 30; $i += ++$j) {
$j = getResult($j);
}
Post-incrementing or pre-incrementing $j directly affects the value of $i according to the examples above. In this case, you need to choose exactly what you want to do.
$i = 0;
echo $i++;
echo $i;
$j=0;
echo ++$j;
echo $j;
Pre increment display incremented value. But Post increment display value then increment. About code will output 01 and 11
What's the difference between ++$i and $i++ in PHP?
++$i is pre-increment whilst $i++ post-increment.
pre-increment: increment variable i first and then de-reference.
post-increment: de-reference and then increment i
"Take advantage of the fact that PHP
allows you to post-increment ($i++)
and pre-increment (++$i). The meaning
is the same as long as you are not
writing anything like $j = $i++,
however pre-incrementing is almost 10%
faster, which means that you should
switch from post- to pre-incrementing
when you have the opportunity,
especially in tight loops and
especially if you're pedantic about
micro-optimisations!"
- TuxRadar
For further clarification, post-incrementation in PHP has been documented as storing a temporary variable which attributes to this 10% overhead vs. pre-incrementation.
++$i increments $i, but evaluates to the value of $i+1
$i++ increments $i, but evaluates to the old value of $i.
Here's an example:
$i = 10;
$a = $i++;
// Now $a is 10, and $i is 11
$i = 10;
$a = ++$i;
// Now $a is 11, and $i is 11
There is sometimes a slight preformance cost for using $i++. See, when you do something like
$a = $i++;
You're really doing this:
$temporary_variable = $i;
$i=$i+1;
$a=$temporary_variable;
++$i is pre-incrementation
$i is incremented
the new value is returned
$i++ is post-incrementation
the value of $i copied to an internal temporary variable
$i is incremented
the internal copy of the old value of $i is returned
++$i //first increment $i then run line
$i++ //first run line then increment $i
this example elplains simply
<?php
$x = 10;
echo $x++. ' '.$x; // the result is 10 and 11
echo '<br>';
$y = 10;
echo ++$y. ' ' .$y; // the result is 11 and 11
// so the $x++ is not showing +1 at first but the next time
// and the ++y is showing +1 first time but not increasing next
in this case there is no difference:
for($i = 0;$i<3;++$i)var_dump $i;
/*
int(0)
int(1)
int(2)
*/
for($i = 0;$i<3;$i++)var_dump $i;
/*
int(0)
int(1)
int(2)
*/
but:
for($i = 0;$i<3; $j = ++$i )var_dump($j);
/*
NULL
int(1)
int(2)
*/
for($i = 0;$i<3; $j = $i++ )var_dump($j);
/*
NULL
int(0)
int(1)
*/
Difference is: ++$i will increment $i variable and return updated value, while $i++ will return original value, so increment it.
$prefix = 1;
$postfix = 1;
echo ++$prefix; // 2
echo $postfix++; // 1
To explain jldupont's point:
$i = 1;
$x = $i++;
echo $x; // prints 1
$x = ++$i;
echo $x; // prints 3
Another way of looking at pre and post incrementing is that it's shorthand for combining 2 statements.
Pre-incrementing
// long form
$y = $y + 1;
$x = $y; // any statement using $y
// shorthand
$x = ++$y; // the same statement using $y
Post-incrementing
// long form
$x = $y; // any statement using $y
$y = $y + 1;
// shorthand
$x = $y++; // the same statement using $y
$i++ is known as post-increment. It increments the value of $i only after assigning the original value of $i to $j first.
++$i is known as pre-increment. It increments the value of $i before assigning the value to $j, so the updated value of $i will be assigned to $j.
Hence,
$i = 4;
$j = $i++;
// Now, $i = 5 and $j = 4
$i = 4;
$j = ++$i;
// Now, $i = 5 and $j = 5
These theories apply in a similar manner for decrementing as well.
Hope this helps!
It's probably best-illustrated by an example...
Post-increment:
$zero = 0;
$n = $zero++; //$n is zero
Pre-increment:
$zero = 0;
$n = ++$zero; //$n is one
Short answer:
Prefix increases the value and returns the value increased
Postfix increases the value and returns the value before it was increased
Prefix is faster
Long answer: If you think a little about it, how you would implement those yourself, you will probably realize why prefix is faster. Truth to be told, postfix is actually (often) implemented using prefix:
const T T::operator ++ (int) // postfix
{
T orig(*this);
++(*this); // call prefix operator
return (orig);
}
Avoid postfix unless you have a specific reason not to. The difference in speed can be quite a lot for complex datatypes.
I actually looked this up a few days ago. Heres my source.
The main purpose of the post-fix increment operator is usage like this:
while(*condition*)
$array[$i++] = $something;
This is a very elegant way, how to get around some array iterations.
Breakdown:
Variable $something will be assigned to the array element indexed with $i
Variable $i will be incremented
Iteration is at the end, condition will be checked
In all other cases, you should use the prefix operator. It makes the code much more clear (You can be sure, that you already work with the incremented value of particular variable).
I ran the following code to test if ++$i is 10% faster than $i++. I admit, the code does not have a stable outcome but even then I should at least have seen some numbers near the 10%. The highest I got was 4-4.5% approximately.
<?php
$randomFloat = rand(0, 10) / 10;
$before1 = microtime(true);
for($i=0; $i <1000000; ++$i){
$rand = (rand(0, 10) / 10) * (rand(0, 10) / 10);
}
$after1 = microtime(true);
echo 'it took '.($after1-$before1) . ' seconds fot ++$i<br />';
$before2 = microtime(true);
for($i=0; $i <1000000; $i++){
$rand = (rand(0, 10) / 10) * (rand(0, 10) / 10);
}
$after2 = microtime(true);
echo 'it took '.($after2-$before2) . ' seconds fot $i++<br /><br />';
echo '++$i is '.((($after1-$before1)*100)/($after2-$before2)-100).'% faster than $i++';
Both operators still do what their syntax implies: to increment. Regardless of prefix or postfix, the variable is sure to be incremented by 1. The difference between the two lies in their return values.
1. The prefix increment returns the value of a variable after it has been incremented.
2. On the other hand, the more commonly used postfix increment returns the value of a variable before it has been incremented.
// Prefix increment
let prefix = 1;
console.log(++prefix); // 2
console.log(prefix); // 2
// Postfix increment
let postfix = 1;
console.log(postfix++); // 1
console.log(postfix); // 2
To remember this rule, I think about the syntax of the two. When one types in the prefix increment, one says ++x. The position of the ++ is important here. Saying ++x means to increment (++) first then return the value of x, thus we have ++x. The postfix increment works conversely. Saying x++ means to return the value of x first then increment (++) it after, thus x++.