Passing arguments to a function by value
In PHP, function arguments are typically passed by value by default. When you pass an argument to a function by value, a copy of the argument’s value is created within the function’s scope. Any changes made to the argument inside the function do not affect the original variable outside the function. Here’s how you pass arguments to a function by value:
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function modifyValue($param) { $param = $param + 10; } $value = 5; modifyValue($value); echo $value; // Outputs: 5
In the example above, the function modifyValue takes an argument $param. When you call modifyValue($value), a copy of the value 5 is created inside the function, and it’s manipulated within the function (increased by 10). However, this change does not affect the original $value variable outside the function, which remains 5.
If you want to modify the original variable inside the function, you can pass it by reference using the & symbol in both the function definition and the function call:
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function modifyValueByReference(&$param) { $param = $param + 10; } $value = 5; modifyValueByReference($value); echo $value; // Outputs: 15
In this case, when you pass $value by reference to modifyValueByReference, any changes made to $param inside the function will directly affect the original $value variable, so it becomes 15 after the function call.
Passing by reference should be used judiciously because it can lead to unexpected behavior if not managed carefully. It’s generally recommended to use pass-by-value for most function arguments and pass-by-reference only when necessary to modify the original variable within the function.