# "((1 + 2 + \$var1) * \$var2) / 2 - 5", ends "7"

• Jan 19th 2011, 11:35 PM
simy202
"((1 + 2 + \$var1) * \$var2) / 2 - 5", ends "7"
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 Math 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
• Jan 21st 2011, 08:29 AM
stebko
If you need answer -8, you forgot ( )
it have to look :: "((1 + 2 + \$var1) * \$var2) / (2 - 5)", ends "7"
without (2-5) it looks like at first divided to 2 and after that minus 5
• Jan 21st 2011, 09:36 AM
CaptainBlack
Quote:

Originally Posted by simy202
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.

((1 + 2 + \$var1) * \$var2) / 2 - 5 evaluates as:

((1 + 2 + 3) * 4) / 2 - 5 the innermost bracket is evaluated first to give (6*4)/2-5, now the multiplications and division take precedence so this becomes 12-5 which is 7.

CB