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The following PHP code will output 3.
function main() { if (1) { $i = 3; } echo $i; } main(); But the following C code will raise a compile error.
void main() { if (1) { int i = 3; } printf("%d", i); } So variables in PHP are not strictly block-scoped? In PHP, variables defined in inner block can be used in outer block?
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PHP has three types of variable scopes: Local variable. Global variable. Static variable.
A local scope is a restricted boundary of a variable within which code block it is declared. That block can be a function, class or any conditional span. The variable within this limited local scope is known as the local variable of that specific code block. The following code block shows a PHP function.
In PHP, variables can be declared anywhere in the script. The scope of a variable is the part of the script where the variable can be referenced/used. PHP has three different variable scopes: local.
There are mainly two types of variable scopes: Local Variables. Global Variables.
PHP only has function scope - control structures such as if don't introduce a new scope. However, it also doesn't mind if you use variables you haven't declared. $i won't exist outside of main() or if the if statement fails, but you can still freely echo it.
If you have PHP's error_reporting set to include notices, it will emit an E_NOTICE error at runtime if you try to use a variable which hasn't been defined. So if you had:
function main() { if (rand(0,1) == 0) { $i = 3; } echo $i; }The code would run fine, but some executions will echo '3' (when the if succeeds), and some will raise an E_NOTICE and echo nothing, as $i won't be defined in the scope of the echo statement.
Outside of the function, $i will never be defined (because the function has a different scope).
For more info: http://php.net/manual/en/language.variables.scope.php
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