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1

I am working on converting many C programs from Unix to Linux and this free() syntax caught my attention:

free((char *) area );

What's the difference between this and free( area ); ?

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3 Answers 3

up vote 12 down vote

The signature of free is:

void free(void *ptr);

So if area is a pointer type that is returned by malloc or its cousins, the conversion in free((char *) area ); is pointless.

Unless... if the code is really, really old, that is, before ANSI C introduced void * as generic pointer. In that ancient time, char * is used as generic pointer.

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up vote 10 down vote

If someone were storing a pointer to dynamic allocated data in a non-pointer type sanctioned by the standard, i.e. area is declared as intptr_t for example, this casts the non-pointer to a pointer type and allows it to be freed.

If area is already a pointer type, this makes no sense at all with any C code written in the last quarter-century.

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up vote 1 down vote

The signature of free() in the standard is wrong. It should have been void free(const void *); to be really correct. This has the consequence that you have to upcast a const * when you want to free it without a warning.

There is a Linus Torvald rant about on why free() should take a const *.

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I'm not convinced about that. malloc returns non-const, so for one thing you need to have a cast in between the malloc and free for this to be needed. Second, const indicates a contract that means you do not modify the contents of the memory pointed to. freeing that memory is a violation of that contract. I can't think think of any arguments for it, so I'd be interested in arguments that are pro-const. –  MicroVirus 8 hours ago
    
See yarchive.net/comp/const.html for the Linus Torvalds's rant. –  hmp 5 hours ago
    
MicroVirus, read Linus' rant, he explains it better as I ever could and he is right. –  tristopia 5 hours ago
    
@MicroVirus: Free doesn't modify the contents. It obliterates its very existence. –  Zan Lynx 2 hours ago

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