What will be the output of the following PHP code?
<?php
$op2 = "blabla";
function foo($op1)
{
echo $op1;
echo $op2;
}
foo("hello");
?>
<?php
$op2 = "blabla";
function foo($op1)
{
echo $op1;
echo $op2;
}
foo("hello");
?>
A. helloblabla
B. Error
C. hello
D. helloblablablabla
Answer: Option B
Solution (By Examveda Team)
The correct answer is Option B: ErrorHere's why:
In PHP, variables defined outside of a function are not automatically accessible inside the function.
Scope is a crucial concept in programming.
The variable `$op2` is defined outside the function `foo()`.
Inside `foo()`, you are trying to use `$op2` without it being defined within the function's scope.
Therefore, PHP will throw an error (specifically, a notice about an undefined variable) when trying to access `$op2` within the function.
Although the `echo $op1` will work fine , but after that when trying to access the `$op2` that is not defined inside the function scope it will throw error .
To make `$op2` accessible inside the function, you would need to either:
1. Pass it as an argument to the function.
2. Use the `global` keyword to bring it into the function's scope.
Since neither of these is done in the given code, an error occurs.
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Related Questions on Functions
Which one of the following is the right way of defining a function in PHP?
A. function { function body }
B. data type functionName(parameters) { function body }
C. functionName(parameters) { function body }
D. function fumctionName(parameters) { function body }
A. Call By Value
B. Call By Reference
C. Default Argument Value
D. Type Hinting
this code will throw an error because the variable $op2 is defined outside the function foo(), and it is not accessible inside the function due to variable scope rules in PHP.