Why Does Returning a Reference to a Local Variable Sometimes Seem to Work?

Returning Reference to Local or Temporary Variable

In the code snippet provided, a function foo() returns a reference to a local variable i. Despite the apparent violation of returning a reference to a variable that no longer exists on the stack, the code does indeed assign the value 6 to the variable i in the main() function.

This is due to a quirk in the memory management of function calls. When a function is invoked, a stack frame is allocated to store the function's local variables. However, even after the function returns, the stack frame remains allocated for a brief period while the function caller's stack frame is still active.

During this time, references to the local variables in the returned function still point to their original memory locations on the stack. This allows the returned reference to be used to access and modify the corresponding variable in the caller's scope.

However, it's crucial to note that this behavior is highly unpredictable and implementation-specific. It may vary depending on the compiler, optimization settings, and runtime environment. Thus, relying on this behavior is generally not advisable and is not considered good programming practice.

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