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Does free(ptr) where ptr is NULL corrupt memory?
I'm writing a C function that frees a pointer if it was malloc()
ed. The pointer can either be NULL (in the case that an error occured and the code didn't get the chance to allocate anything) or allocated with malloc()
. Is it safe to use free(ptr);
instead of if (ptr != NULL) free(ptr);
?
gcc
doesn't complain at all, even with -Wall -Wextra -ansi -pedantic开发者_运维问答
, but is it good practice?
Quoting the C standard, 7.20.3.2/2 from ISO-IEC 9899:
void free(void *ptr);
If
ptr
is a null pointer, no action occurs.
Don't check for NULL
, it only adds more dummy code to read and is thus a bad practice.
However, you must always check for NULL
pointers when using malloc
& co. In that case NULL
mean that something went wrong, most likely that no memory was available.
It is good practice to not bother checking for NULL
before calling free
. Checking just adds unnecessary clutter to your code, and free(NULL)
is guaranteed to be safe. From section 7.20.3.2/2 of the C99 standard:
The
free
function causes the space pointed to byptr
to be deallocated, that is, made available for further allocation. Ifptr
is a null pointer, no action occurs.
As noted in the comments, some people sometimes wonder if checking for NULL
is more efficient than making a possibly unnecessary function call. However, this:
- Is a premature micro-optimization.
- Shouldn't matter. Checking for
NULL
first even might be a pessimization. For example, if 99% of the time your pointers aren'tNULL
, then there would be a redundantNULL
check 99% of the time to avoid an extra function call 1% of the time.
See http://linux.die.net/man/3/free which states:
If ptr is NULL, no operation is performed.
In my opinion, no, at least not in your case.
If you couldn't allocate memory, you should have checked that WAY before the call of free.
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