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PHP Beginnings Ex. #4: Arithmetic-Assignment Operators and Variables

Arithmetic-assignment operators perform an arithmetic operation on the variable at the same time as assigning a new value. For this PHP exercise, write a script to reproduce the output below. Manipulate only one variable using no simple arithmetic operators to produce the values given in the statements.

Hint: In the script each statement ends with "Value is now $variable."

Value is now 8.
Add 2. Value is now 10.
Subtract 4. Value is now 6.
Multiply by 5. Value is now 30.
Divide by 3. Value is now 10.
Increment value by one. Value is now 11.
Decrement value by one. Value is now 10.

 

 

Here's the script:

  1. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"     
  2. "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">   
  3. <html xmlns="http://www.w3.org/1999/xhtml"  xml:lang="en" lang="en"> 
  4. <head> 
  5. <meta http-equiv="content-type" content="text/html;charset=iso-8859-1" /> 
  6. <title>Arithmetic-Assignment Operators</title> 
  7. </head>   
  8.  
  9. <body>   
  10.  
  11. <?php   
  12.  
  13. $num = 8; 
  14. echo "Value is now $num.<br/>";   
  15.  
  16. $num += 2; 
  17. echo "Add 2. Value is now $num. <br/>";   
  18.  
  19. $num -= 4; 
  20. echo "Subtract 4. Value is now $num. <br/>";   
  21.  
  22. $num *= 5; 
  23. echo "Multiply by 5. Value is now $num. <br/>";   
  24.  
  25. $num /= 3; 
  26. echo "Divide by 3. Value is now $num. <br/>";   
  27.  
  28. $num++; 
  29. echo "Increment value by one. Value is now $num.<br/>";   
  30.  
  31. $num--; 
  32. echo "Decrement value by one. Value is now $num.";   
  33.  
  34. ?> 
  35.  
  36. </body> 
  37. </html>

See the output of the script in a separate window here. You can also view the output's HTML source from the new window, if you need to check that.

To open a PHP code editor in a new tab, click here.

Comments

Submitted by tenlee on Tue, 06/12/2012 - 23:05.

Please help me solve this..
its look kind a weird when I tried to echoing the $num like this :

note : Im using $result as $num

echo 'Value now is '. $result = 8 . '<br/>';
echo 'Add 2, Value now is '. $result +=2; echo '<br/>';
echo 'Subtract 4, Value now is '. $result -= 4; echo '<br/>';
echo 'Multiply by 5, Value now is '. $result *= 5; echo '<br/>';
echo 'Divide by 3, Value now is '. $result /= 3; echo '<br/>';
echo 'Increment by one, Value now is '. $result++ ;echo '<br/>';
echo 'Decrement by one, Value now is '. $result-- ; echo '<br/>';

and the output is like this:

Value now is 8
Add 2, Value now is 10
Subtract 4, Value now is 6
Multiply by 5, Value now is 30
Divide by 3, Value now is 10
Increment by one, Value now is 10
Decrement by one, Value now is 11

Why the value in the last value (increment and decrement by one) is miscalculated?
Am I missing something?

I appreciate all of your effort for helping me,
thanks before.. :)

Submitted by ebricca on Wed, 06/13/2012 - 10:38.

you are actually using the direct return value from the decrement function.
if you write
echo "mytext".$result--

the correct return result would be given by
echo "mytext".(--$result)

another problem that you noticed was that the Dot (.) operator is not always processed last
so braces () become mandatory.

so i guess what you wanted to accomplish was

echo 'Value now is '.($result = 8).'';
echo 'Add 2, Value now is '.($result +=2).'';
echo 'Subtract 4, Value now is '.($result -= 4).'';
echo 'Multiply by 5, Value now is '.($result *= 5).'';
echo 'Divide by 3, Value now is '.($result /= 3).'';
echo 'Increment by one, Value now is '.(++$result).'';
echo 'Decrement by one, Value now is '.(--$result).'';

Submitted by arman27 on Sun, 09/02/2012 - 03:25.

In $result++, the result is first stored and then incremented and that same for $result--.

After the division of the value in $result by three, quotient is stored in $result. When $result++ is echoed it will show the value currently present there i.e.,10. and not 11 because still now it is not incremented. Later on the value in $result is incremented and that's shown while decrement operator is used. Only after that value in $result will be decremented.

If again the value of $result is echoed after that result will be 10. That's because now the $result contain the decremented value.

Whereas if ++$result or --$result is used then first the value in $result will be incremented or decremented respectively and then it will be stored in $result.

Thank you.

Submitted by EmilShamloo on Fri, 09/29/2017 - 03:31.

Hi, if you add your -- or ++ before your variable, it will print correct.
Search this page for:
PHP Increment / Decrement Operators
https://www.w3schools.com/PhP/php_operators.asp

Submitted by Nas on Wed, 09/26/2012 - 14:28. 
$variable = 8;
echo "Value is now ".$variable."<br />";

$variable += 2;
echo "Add 2. Value is now ".$variable."<br />";

$variable -= 4;
echo "Subtract 4. Value is now ".$variable."<br />";

$variable *= 5;
echo "Multiply by 5. Value is now ".$variable."<br />";

$variable /= 3;
echo "Divide by 3. Value is now ".$variable."<br />";

$variable++;
echo "Increment value by 1. Value is now ".$variable."<br />";

$variable--;
echo "Decrement value by 1. Value is now ".$variable."<br />";

Submitted by boksa on Fri, 12/14/2012 - 10:27. 
$x = 0;
	$x += 8;
	printf ("Value is : %d<br/>",$x);
	$x +=2;
	printf ("Value is : %d<br/>",$x);
	$x -=4;
	printf ("Value is : %d<br/>",$x);
	$x *=5;
	printf ("Value is : %d<br/>",$x);
	$x /=3;
	printf ("Value is : %d<br/>",$x);
	$x++;
	printf ("Value is : %d<br/>",$x);
	$x--;
	printf ("Value is : %d<br/>",$x);

Submitted by kenneth24 on Fri, 05/30/2014 - 21:59. 
<?php

$base = 8;

echo "value is now ".$base."<br>"; 

for($i=0; $i<6; $i++)
{
	
	switch ($i)
	{
	
		case 0;
			
			$base = $base + 2;
			
			echo "Add 2. value is now ".$base."<br>";
						
		break;
		
		case 1;
		
			$base = $base - 4;
			echo "Subtract 4. Value is now ".$base."<br>";
		
		break;
		
		case 2;
			
			$base = $base * 5;
			echo "Multiply 5. Value is now ".$base."<br>";
		
		break;
		
		case 3;
		
			$base = $base / 3;
			echo "Divide by 3. Value is now ".$base."<br>"; 
			
		break;
		
		case 4;
		
			$base = ++$base;
			echo "Increment Value by one. Value is now ".$base."<br>";
			
		break;
		
		case 5;
		
			$base = --$base;
			echo "Decrement value by one. Value is now ".$base."<br>";
			
		break;
	}
}
?>

Submitted by adamk90 on Sat, 06/21/2014 - 14:45. 
 <?php
  $commands = array(
	'add' => "+",
	'substract' => "-",
	'multiply' => "*",
	'divide' => "/",
	'increment' => "+",
	'decrement' => "-");

  $thevalue = 8;

  function docommand($command) {
	global $commands, $thevalue;
	$parse_command = explode(' ', $command);
        if ($commands[@strtolower($parse_command[0])] != NULL) {
		$dowith = strtolower($parse_command[0]);
		$value = $parse_command[count($parse_command)-1];
		$result = "$command. " . eval("\$thevalue = $thevalue $commands[$dowith] $value;") . "Value is now $thevalue.";
		echo "$result<br>";
	} else {
		echo "Invalid command: $parse_command[0]<br>";
	}
  }
  
  echo "Value is now $thevalue. <br>";
  docommand("Add 2");
  docommand("Increment value by 6");
  docommand("Decrement value by 8");
  docommand("Multilpy value by 10");
  docommand("Divide by 5");
  docommand("Multiply this $thevalue with $thevalue");
 ?>

Submitted by ritho on Sat, 07/26/2014 - 10:09. 

$s= "Value is now ";
$v = 0;
echo  $s . ($v+=8) .'.';
echo "\nAdd2. " . $s . ($v+=2) .'.';
echo "\nSubtract 4. " . $s . ($v-=4) .'.'; 
echo "\nMultiply by 5. " . $s . ($v*=5) .'.';
echo "\nDivide by 3. " . $s . ($v/=3) .'.';
echo "\nIncrement value by one. " . $s . ++$v .'.';
echo "\nDecrement value by one. " . $s . --$v .'.';

Submitted by seant on Wed, 12/17/2014 - 07:30.

this was the result when

Value is now 8Add 2. Value is now 8
Subtract 4. Value is now 8
Multiply by 5. Value is now 8
Divide by 30. Value is now 8
Increment value by 1. Value is now 8
Decrement value by 1. Value is now 8

when I put a variable in a string and updated the variable

<?php 
$variable = 8;
$a = "Value is now $variable";
echo " $a";
$variable += 2 ;
echo "Add 2. $a". "<br />";
$variable -= 4 ;
echo " Subtract 4. $a" . "<br />";
$variable *= 5 ;
echo " Multiply by 5. $a" . "<br />";
$variable /= 30 ;
echo " Divide by 30. $a" . "<br />";
$variable += 1 ;
echo " Increment value by 1. $a" . "<br />";
$variable -= 1 ;
echo " Decrement value by 1. $a" . "<br />";
?>

Submitted by seant on Wed, 12/17/2014 - 07:25.

i put a variable in a string. So if the value of variable changes shouldn't the string get updated by itself
because this code 

<?php 
$variable = 8;
$a = "Value is now $variable";
echo " $a";
$variable += 2 ;
echo "Add 2. $a". "<br />";
$variable -= 4 ;
echo " Subtract 4. $a" . "<br />";
$variable *= 5 ;
echo " Multiply by 5. $a" . "<br />";
$variable /= 30 ;
echo " Divide by 30. $a" . "<br />";
$variable += 1 ;
echo " Increment value by 1. $a" . "<br />";
$variable -= 1 ;
echo " Decrement value by 1. $a" . "<br />";
?>

Gave the following result

Value is now 8Add 2. Value is now 8
Subtract 4. Value is now 8
Multiply by 5. Value is now 8
Divide by 30. Value is now 8
Increment value by 1. Value is now 8
Decrement value by 1. Value is now 8

Submitted by pequenosaltamontes on Tue, 03/03/2015 - 14:58.

I had some trouble by using the following statement for increment and decrement:
print "Increment value by one. Value is now " . $var++. "";
print "Decrement value by one. Value is now " . $var--. "";

So I decided to do the $varr++ and $var-- before the print. Here is my solution. 10 lines of a simple solution, no fancy arrays, no luxurious loops :)

            $var=8;
            print "Value is now " . $var . "<br>";
            print "Add 2. Value is now " . ($var+=2) . "<br>";
            print "Substract 4. Value is now " . ($var-=4) . "<br>";
            print "Multiply by 5. Value is now " . ($var*=5) . "<br>";
            print "Divide by 3. Value is now " . ($var/=3) . "<br>";
            $var++;
            print "Increment value by one. Value is now " . $var . "<br>";
            $var--;
            print "Decrement value by one. Value is now " . $var . "<br>";

Submitted by EmilShamloo on Fri, 09/29/2017 - 03:46.

Hi,
Try this one:

  1. $var=8;
  2. print "Value is now " . $var . "<br>";
  3. print "Add 2. Value is now " . ($var+=2) . "<br>";
  4. print "Substract 4. Value is now " . ($var-=4) . "<br>";
  5. print "Multiply by 5. Value is now " . ($var*=5) . "<br>";
  6. print "Divide by 3. Value is now " . ($var/=3) . "<br>";
  7. print "Increment value by one. Value is now". (++$var). "<br>";
  8. // if ++ comes after, it is not working!
  9. print "Decrement value by one. Value is now". (--$var). "<br>";
  10. // When -- is written after your variable, then it will work.