C language. free() and changed pointer

Question

Will free() work correctly if I change pointer the following way?

ar_byte_100=calloc(100,1);
ar_byte_100=&((unsigned long int*)ar_byte_100)[50];
free(ar_byte_100);

Nope(or might as it invokes UB). You have to give `free` the pointer returned by `calloc`. — Spikatrix, Apr 08 '15 at 09:39
A similar question, regarding C++ rather than C: http://stackoverflow.com/q/29192430/4200092 — GoBusto, Apr 08 '15 at 09:40
@GoBusto IMHO, No need to go for `C++` here, lots of `C` dupes are already there. :-) — Sourav Ghosh, Apr 08 '15 at 09:43

score 2 · Answer 1 · answered Apr 08 '15 at 09:39

2

No.

You must pass the exact pointer returned by malloc() or calloc() to free().

answered Apr 08 '15 at 09:39

Jonathon Reinhart

score 2 · Answer 2 · answered Apr 08 '15 at 09:40

2

According to the standard free(ptr)

ptr should be a pointer returned by malloc() calloc() and realloc() anything other than that is a undefined behavior.

answered Apr 08 '15 at 09:40

Gopi

And can I `realloc(ar_byte_100, 0); ar_byte_100=&((unsigned long int*)ar_byte_100)[-50];realloc(ar_byte_100, 0);`? – user4753163 Apr 08 '15 at 11:59

2 Answers2