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## Topic review (newest first)

gAr
2011-02-09 13:38:21

Hi Shivamcoder3013,

PHP code doesn't work outside <?php   ?> tags. It will simply print it as it is.

ShivamS
2011-02-09 09:36:45

I am quite confused. How can you expect the PHP code to work if it is outside <?php and ?> ?

gAr
2011-01-18 17:51:34

Hi simy202,

I guess you haven't read about operator precedence.
Please refer this: http://www.php.net/manual/en/language.operators.precedence.php

echo((1  + 2 + \$var1) * \$var2) / 2 - 5;

This evaluates to : (((1 + 2 + 3)*4)/2) - 5 = ((6*4)/2) - 5 = 12 - 5 = 7.
If you want it to be -8, rewrite it as:

echo((1  + 2 + \$var1) * \$var2) / (2 - 5);

The left of '=' sign is called Left hand side, not the left of '/' sign...

simy202
2011-01-18 17:14:33

I'm going through a php tutorial right now and this math example below is driving me nuts. The exmaple was "((1 + 2 + \$var1) * \$var2) / 2 - 5", and all the rest of the code below is just me trying to figure it out.

The answer ends up being "7", but to me it looks like PHP is sayding "24 / -3 = 7" when it should be "24 / -3 = -8". What am I missing in this PHP Math issue?? Thanks!

#### Code:

```<?php
\$var1 = 3;
\$var2 = 4;
?>
\$var1 = 3;<br />
\$var2 = 4;<br /><br />
<br />
<h1> ((1  + 2 + \$var1) * \$var2) / 2 - 5 = <?php echo((1  + 2 + \$var1) * \$var2) / 2 - 5; ?><br /> </h1>
<hr />
<br />
<br />
LEFT SIDE:<br />
(1  + 2 + \$var1) * \$var2 =  <?php echo ((1  + 2 + \$var1) * \$var2); ?><br />
<br />
RIGHT SIDE:<br />
2 - 5 =   <?php echo 2 - 5; ?><br />
<br />
24 / -3 SHOULD BE:
<h1>24 / -3 = <?php echo 24 / -3; ?><br /></h1>```

That spits out:

\$var1 = 3;
\$var2 = 4;

((1 + 2 + \$var1) * \$var2) / 2 - 5 = 7

LEFT SIDE:
(1 + 2 + \$var1) * \$var2 = 24

RIGHT SIDE:
2 - 5 = -3

24 / -3 SHOULD BE:
24 / -3 = -8