Php Variable Assignment

Php Variable Assignment-4
After the division of the value in $result by three, quotient is stored in $result. so i guess what you wanted to accomplish was echo 'Value now is '.($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).''; In $result , the result is first stored and then incremented and that same for $result--.

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Hint: In the script each statement ends with "Value is now $variable." Value is now 8. If again the value of $result is echoed after that result will be 10.

When $result is echoed it will show the value currently present there i.e.,10.

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

Value is now ".$variable.""; $variable ; echo "Increment value by 1.

Value is now ".$variable.""; $variable--; echo "Decrement value by 1.

Value is now ".$variable.""; $x = 0; $x = 8; printf ("Value is : %d",$x); $x =2; printf ("Value is : %d",$x); $x -=4; printf ("Value is : %d",$x); $x *=5; printf ("Value is : %d",$x); $x /=3; printf ("Value is : %d",$x); $x ; printf ("Value is : %d",$x); $x--; printf ("Value is : %d",$x); "-"); $thevalue = 8; function docommand($command) echo "Value is now $thevalue. Literal strings do not attempt to parse special characters or variables.If using single quotes, you could enter a variable name into a string like so: When using interpolation, it is often the case that the variable will be touching another character.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. $variable = 8; echo "Value is now ".$variable.""; $variable = 2; echo "Add 2. That said, there are a few different types of strings and they offer slightly different syntax, with slightly different behaviors.Single quotes are used to denote a “literal string”. $result /= 3; echo ''; echo 'Increment by one, Value now is '. $result *= 5; echo ''; echo 'Divide by 3, Value now is '. With more than a decade's professional experience, he is a published author, a frequent blogger and speaker, and an outspoken advocate of standards-based development.Arithmetic-assignment operators perform an arithmetic operation on the variable at the same time as assigning a new value. and not 11 because still now it is not incremented.


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