How does free know how much to free?
In C programming, you can pass any kind of pointer you like as an argument to free, how does it know the size of the allocated memory to free? Whenever I pass a pointer to some function, I have to also pass the size (ie an array of 10 elements needs to receive 10 as a parameter to know the size of the array), but I do not have to pass the size to the free function. Why not, and can I use this same technique in my own functions to save me from needing to cart around the extra variable of the array's length?
When you call malloc() , you specify the amount of memory to allocate. The amount of memory actually used is slightly more than this, and includes extra information that records (at least) how big the block is. You can't (reliably) access that other information - and nor should you :-).
When you call free() , it simply looks at the extra information to find out how big the block is.
Most implementations of C memory allocation functions will store accounting information for each block, either in-line or separately.
One typical way (in-line) is to actually allocate both a header and the memory you asked for, padded out to some minimum size. So for example, if you asked for 20 bytes, the system may allocate a 48-byte block:
The address then given to you is the address of the data area. Then, when you free the block, free will simply take the address you give it and, assuming you haven't stuffed up that address or the memory around it, check the accounting information immediately before it. Graphically, that would be along the lines of:
____ The allocated block ____
/
+--------+--------------------+
| Header | Your data area ... |
+--------+--------------------+
^
|
+-- The address you are given
Keep in mind the size of the header and the padding are totally implementation defined (actually, the entire thing is implementation-defined (a) but the in-line accounting option is a common one).
The checksums and special markers that exist in the accounting information are often the cause of errors like "Memory arena corrupted" or "Double free" if you overwrite them or free them twice.
The padding (to make allocation more efficient) is why you can sometimes write a little bit beyond the end of your requested space without causing problems (still, don't do that, it's undefined behaviour and, just because it works sometimes, doesn't mean it's okay to do it).
(a) I've written implementations of malloc in embedded systems where you got 128 bytes no matter what you asked for (that was the size of the largest structure in the system), assuming you asked for 128 bytes or less (requests for more would be met with a NULL return value). A very simple bit-mask (ie, not in-line) was used to decide whether a 128-byte chunk was allocated or not.
Others I've developed had different pools for 16-byte chunks, 64-bytes chunks, 256-byte chunks and 1K chunks, again using a bit-mask to decide what blocks were used or available.
Both these options managed to reduce the overhead of the accounting information and to increase the speed of malloc and free (no need to coalesce adjacent blocks when freeing), particularly important in the environment we were working in.
From the comp.lang.c FAQ list: How does free know how many bytes to free?
The malloc/free implementation remembers the size of each block as it is allocated, so it is not necessary to remind it of the size when freeing. (Typically, the size is stored adjacent to the allocated block, which is why things usually break badly if the bounds of the allocated block are even slightly overstepped)
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