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armanriazi•rust•mem•doublefree

This is a problem: when s2 and s1 (s2 is copied s1 means 2different pointer and the same data) go out of scope, they will both try to free the same memory. This is known as a double free error and is one of the memory safety bugs we mentioned previously. Freeing memory twice can lead to memory corruption, which can potentially lead to security vulnerabilities.

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