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When a variable is declared within a block of code, its scope defines the portions of the code where it can be accessed. But what happens to the memory allocated to the variable when its scope ends? Is it overwritten immediately, or does it remain reserved until the function in which it resides concludes?
This question arises from the following code snippet:
foo() { int *p; { int x = 5; p = &x; } int y = *p; // undefined behavior }
If the memory of the variable x is still allocated after it goes out of scope, the pointer p will continue to point to the correct location, allowing us to access the value of x even though it is no longer in scope.
To unravel this enigma, we must delve into the concepts of scope and lifetime.
Scope encompasses the sections of code where a variable can be accessed. When a variable is declared within curly braces ({}), its scope is limited to the enclosed code block.
Lifetime encompasses the duration where a variable or object exists in a valid state. For automatic or local non-static variables, their lifetime is confined to their scope. In other words, these variables are destroyed automatically once their scope ends.
In the given code snippet, the variable x is declared as a non-static local variable, meaning its lifetime is bound to its scope. When the inner scope ends, x ceases to exist, leaving the pointer p pointing to a location that is no longer valid.
As a result, attempting to access the value of x through p is undefined behavior. The compiler may or may not allocate memory to x beyond its scope, leading to unpredictable outcomes. It is therefore crucial to avoid accessing variables whose scope has ended.
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