CODEWITHSUNDEEP
php
user2533777
user2533777

Reputation:

unexpected output in php while using `!`

I was just trying an small program but i am getting unexpected output.

for($i=20;!$i<20;$i--)
echo '*';

the expected output is * as only for first case when $i=20 is false so !$i<20 should return true but the no of times loop is executing is equal to value of $i.

I tried manipulating values and concluded when I set value of $i in negative the loop gets infinite.

Further i tried this 

echo 20<20;

output was nothing as expected then 

echo !20<20;

output was 1 as expected

Now when it tried:

19<20

it is returning 1 but when i am trying

!19<20

it is returning 1 why did this occurred ?? I am running PHP on WAMP Server and my PHP Version 5.5.0

Note: I am not having any problem with for loop i can handle it so please don't answer correcting my loop rather I was confused with working of ! so please answer for it.

Upvotes: 0

Views: 49

Answers (2)

Vijay Verma
Vijay Verma

Reputation: 3698

Try 

for($i=20;!($i<20);$i--)
echo '*';

Problem was that '!$i' were executing first and then '<' operator work.

Upvotes: 0

Josh B
Josh B

Reputation: 1868

You need brackets to 'not' the right part:

for($i=20;!($i<20);$i--)
  echo '*';

The example of !20<20 does this:

!20<20
!(true)<20  <- converts the type to bool so we can negate
false<20    <- negates the true to false
0<20        <- converts the false to an int to compare
true

And the example of !19<20 does this:

!19<20
!(true)<20  <- converts the type to bool so we can negate
false<20    <- negates the true to false
0<20        <- converts the false to an int to compare
true

Upvotes: 1

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