Operators And Expressions In Php — Part 2 MCQs with Explanations

There will no output returned as the variable $red does not hold any value.

You can use numbers in a variable name.

A variable name can not start with a numeric value.

Unlike other programming languages there are no data types in PHP.

You can not use var before a variable name.

Since there is a semi-colon in between $hello and $bye, the line ends at $hello. However $bye would have printed if a echo was present before “$bye”.

Since the variable x is not initialised it is not storing any value, therefore nothing will be printed on the screen.

Varibale x stores the value 10 and not 5.

The variable x is not defined inside the function fun(), therefore nothing is printed on the screen.

First when the function is called variable x is initialised to 10 so 10 is printed later the global value 5 is printed.

Firstly, the statements outside the function are printed, since z is not defined it’ll no value is printed for z. Next the function is called and the value of z inside the function is printed.

It is same as above but the value passed into the function is 3,4 and not 4,3. Therefore the difference in answer.

Value 3, 4 is passed to the function but that is lost because x and y are initialised to 4 and 3 inside the function. Therefore we get the given result.

In C you won’t get an error but in PHP you’ll get a syntax error.

Understanding the PHP Code: The code defines a function named 'fun' that takes two arguments, $a and $b . Inside the function: The value of $b is changed to 3. The function calculates $a - $b + $b - $a , which simplifies to 0 (because the $a and -a cancel each other out, and $b and -b cancel each other out). The function then returns this result (which is 0). Outside the function: The values of $a and $b are initially set to 10 and 4 respectively. The function 'fun' is called with 10 and 4 as arguments, and the returned value (0) is assigned to $c . Finally, the code prints the values of $a , $b , and $c . Tracing the Output: $a remains 10 (it is not modified within the function because of pass-by-value). $b remains 4 (the change inside the function doesn't affect the global $b ). $c becomes 0 (the return value of the function). Therefore, the code will print "1040". Correct Option: The correct answer is Option A: 1040

Since variables $winner and $looser is not defined we only see / as output.

You can access the global variable using $GLOBALS[‘globalvariablename’].

The value of global variable y does not change therefore it’ll print 10;

Every time the function is called the value of x becomes 0, therefore we get 0 on every function call.

When static is used, each time the function is called, that variable will still have the information it contained from the last time the function was called.

Since variable x is not defined inside the function fun(), nothing will be printed.

Since, we are using $GLOBALS[‘x’], the question is similar to question 7.

This question is similar to question 6.

++$a increments a and then prints it,$a++ prints and then increments.

The + operator does union of arrays in that order, then the === operator compares key and value pairs.

The x .= y is a shorthand for x = x.y and this is evaluated from right to left.

x is preincremented and y is post incremented thus both are 2 in the if condition, later y is incremented.

Expression is evaluated from right to left.

First $y = 2 is compared to and then decremented, then incremented and compared to $y = 2.

The ** is not a valid operator,only ++ and — exist.

–$y == 2 is false but y is decremented, the xor gives true if only one of the operands are true, thus 1 xor 0 is true.

–$y == 2 is false but y is decremented, the xor gives true if only one of the operands are true, thus 1 xor 0 is true.

Automatically x is declared and initialized to 0,then decremented and compared with its increments, thus returns 1.

Result is x&&y which returns 1 if both x and y are true.

Operator precedence order must be followed.

In if condition when the first case is 0 and is an && operation then the second command is not executed.

In if condition when the first case is 1 and is an || operation then the second command is not executed.

The first condition returns true and increments but the second condition is false.

Post increment of a is done after expression evaluation.

Follow the order of post and pre increments.