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Q&A Can freed pointers undergo lvalue conversion?

I see it such that it is undefined behavior (I refer to N1570 as the OP https://port70.net/~nsz/c/c11/n1570.html): After free(p), the lifetime of the object pointed to has ended (7.22.3p1: "The li...

posted 1y ago by Dirk Herrmann‭

Answer
#1: Initial revision by user avatar Dirk Herrmann‭ · 2023-04-13T19:14:30Z (about 1 year ago)
I see it such that it is undefined behavior (I refer to N1570 as the OP https://port70.net/~nsz/c/c11/n1570.html):

After `free(p)`, the lifetime of the object pointed to has ended (7.22.3p1: "The lifetime of an allocated object extends from the allocation until the deallocation.")

Consequently, the value of `p` becomes indeterminate (6.2.4p2: "The value of a pointer becomes indeterminate when the object it points to (or just past) reaches the end of its lifetime.")

Indeterminate means, it could be a trap representation (3.19.2: "indeterminate value[:] either an unspecified value or a trap representation")

Certainly, the `free(p)` does not as such change the value of `p`, but I understand this phrase such that it gives the compiler some freedom for optimization.  For example, the stack bytes used to hold `p` might be re-used for temporary calculations etc.  Such values do not necessarily have to be pointer values, and thus they may have the effect that the value of `p` is changed to a trap representation. 

Accessing `p` after `free(p)` is therefore undefined behavior: 6.2.6.1 p5: "If the stored value of an object has such a representation and is read by an lvalue expression that does not have character type, the behavior is undefined."