I want to test how much the OS does allocate when I request 24M memory.
for (i = 0; i < 1024*1024; i++)
ptr = (char *)malloc(24);
When I write like this I get RES is 32M from the top command.
ptr = (char *)malloc(24*1024*1024);
But when I do a little change the RES is 244. What is the difference between them? Why is the result 244?
The allocator has its own data structures about the bookkeeping that require memory as well. When you allocate in small chunks (the first case), the allocator has to keep a lot of additional data about where each chunk is allocated and how long it is. Moreover, you may get gaps of unused memory in between the chunks because malloc has a requirement to return a sufficiently aligned block, most usually on an 8-byte boundary.
In the second case, the allocator gives you just one contiguous block and does bookkeeping only for that block.
Always be careful with a large number of small allocations, as the bookkeeping memory overhead may even outweigh the amount of the data itself.
The second allocation barely touches the memory. The allocator tells you "okay, you can have it" but if you don't actually touch the memory, the OS never actually gives it to you, hoping you'll never use it. Bit like a Ponzi scheme. On the other hand, the other method writes something (a few bytes at most) to many pages, so the OS is forced to actually give you the memory.
Try this to verify, you should get about 24m usage:
memset(ptr, 1, 1024 * 1024 * 24);
In short, top doesn't tell you how much you allocated, i.e. what you asked from malloc. It tells you what the OS allocated to your process.
In addition to what has been said:
It could be that some compilers notice how you allocate multiple 24 Byte Blocks in a loop, assigning their addresses to the same pointer and keeping only the last block you allocated, effectively rendering every other malloc from before useless. So it may optimize your whole loop into something like this:
ptr = (char *)malloc(24);
i = 1024*1024;
Related
int main() {
int i = 0, ARRAY_SIZE = 500000;
char **char_array;
char_array = (char **)malloc(ARRAY_SIZE * sizeof(char*));
//physical memory used before loop = M KB
for (i = 0; i < ARRAY_SIZE; i++) {
char_array[i] = (char *)malloc(16 * sizeof(char));
}
//physical memory usage after loop = M+19532 KB
return 0;
}
I have the above piece of code. I don't understand where the 19532 KB of memory use is coming from. In my machine (64 bit), sizeof(char*) should be 8 bytes. For row initialization of an array, how is memory used? I'm a beginner in C, so any help would be appreciated.
Every call to malloc uses some additional "overhead" memory beyond the amount you actually request. The library needs some space to keep track of those blocks to know how to free them later, and it may round up the size of small allocations for alignment.
Your allocation of char_array itself needs 500000 * 8, about 4 megabytes of memory, and the overhead is probably negligible. So it looks like your 500000 allocations of 1 byte each are actually using about 32 bytes each. This wouldn't be too unusual; for instance, malloc might be rounding up the size to 16 bytes for alignment, and then needing an additional 16 bytes for bookkeeping (say, 8 bytes each for a size and a pointer to the next block to make a linked list).
Obviously, this is a very inefficient way to allocate space for 500000 characters. You should instead just create an array of 500000 char (not pointers) and index into it directly.
malloc is allowed to arrange and "align" the blocks of memory you allocate in really any way it wishes, and in practice there are both minimum allocation sizes and internal bookkeeping overhead for each block. Single-byte allocations (inside the loop you're allocating chars and not char *s) are uncommon because they're inefficient to do like this; in practice you're probably actually allocating (at least) 8 bytes plus the internal overhead, each.
Whatever "physical memory" measure you're looking at does not map precisely to what you're looking for here. malloc generally takes pools of memory from the OS and then doles it out into the smaller allocated blocks (with internal overhead) that you see at the level of your program. I would not expect any kind of exact correspondence between bytes allocated by the user program via malloc, and what the OS reports about the process. Perhaps a directional correlation holding all else equal, but not something worth spending much time trying to deduce conclusions from.
Surprisingly simple/stupid/basic question, but I have no idea: Suppose I want to return the user of my function a C-string, whose length I do not know at the beginning of the function. I can place only an upper bound on the length at the outset, and, depending on processing, the size may shrink.
The question is, is there anything wrong with allocating enough heap space (the upper bound) and then terminating the string well short of that during processing? i.e. If I stick a '\0' into the middle of the allocated memory, does (a.) free() still work properly, and (b.) does the space after the '\0' become inconsequential? Once '\0' is added, does the memory just get returned, or is it sitting there hogging space until free() is called? Is it generally bad programming style to leave this hanging space there, in order to save some upfront programming time computing the necessary space before calling malloc?
To give this some context, let's say I want to remove consecutive duplicates, like this:
input "Hello oOOOo !!" --> output "Helo oOo !"
... and some code below showing how I'm pre-computing the size resulting from my operation, effectively performing processing twice to get the heap size right.
char* RemoveChains(const char* str)
{
if (str == NULL) {
return NULL;
}
if (strlen(str) == 0) {
char* outstr = (char*)malloc(1);
*outstr = '\0';
return outstr;
}
const char* original = str; // for reuse
char prev = *str++; // [prev][str][str+1]...
unsigned int outlen = 1; // first char auto-counted
// Determine length necessary by mimicking processing
while (*str) {
if (*str != prev) { // new char encountered
++outlen;
prev = *str; // restart chain
}
++str; // step pointer along input
}
// Declare new string to be perfect size
char* outstr = (char*)malloc(outlen + 1);
outstr[outlen] = '\0';
outstr[0] = original[0];
outlen = 1;
// Construct output
prev = *original++;
while (*original) {
if (*original != prev) {
outstr[outlen++] = *original;
prev = *original;
}
++original;
}
return outstr;
}
If I stick a '\0' into the middle of the allocated memory, does
(a.) free() still work properly, and
Yes.
(b.) does the space after the '\0' become inconsequential? Once '\0' is added, does the memory just get returned, or is it sitting there hogging space until free() is called?
Depends. Often, when you allocate large amounts of heap space, the system first allocates virtual address space - as you write to the pages some actual physical memory is assigned to back it (and that may later get swapped out to disk when your OS has virtual memory support). Famously, this distinction between wasteful allocation of virtual address space and actual physical/swap memory allows sparse arrays to be reasonably memory efficient on such OSs.
Now, the granularity of this virtual addressing and paging is in memory page sizes - that might be 4k, 8k, 16k...? Most OSs have a function you can call to find out the page size. So, if you're doing a lot of small allocations then rounding up to page sizes is wasteful, and if you have a limited address space relative to the amount of memory you really need to use then depending on virtual addressing in the way described above won't scale (for example, 4GB RAM with 32-bit addressing). On the other hand, if you have a 64-bit process running with say 32GB of RAM, and are doing relatively few such string allocations, you have an enormous amount of virtual address space to play with and the rounding up to page size won't amount to much.
But - note the difference between writing throughout the buffer then terminating it at some earlier point (in which case the once-written-to memory will have backing memory and could end up in swap) versus having a big buffer in which you only ever write to the first bit then terminate (in which case backing memory is only allocated for the used space rounded up to page size).
It's also worth pointing out that on many operating systems heap memory may not be returned to the Operating System until the process terminates: instead, the malloc/free library notifies the OS when it needs to grow the heap (e.g. using sbrk() on UNIX or VirtualAlloc() on Windows). In that sense, free() memory is free for your process to re-use, but not free for other processes to use. Some Operating Systems do optimise this - for example, using a distinct and independently releasble memory region for very large allocations.
Is it generally bad programming style to leave this hanging space there, in order to save some upfront programming time computing the necessary space before calling malloc?
Again, it depends on how many such allocations you're dealing with. If there are a great many relative to your virtual address space / RAM - you want to explicitly let the memory library know not all the originally requested memory is actually needed using realloc(), or you could even use strdup() to allocate a new block more tightly based on actual needs (then free() the original) - depending on your malloc/free library implementation that might work out better or worse, but very few applications would be significantly affected by any difference.
Sometimes your code may be in a library where you can't guess how many string instances the calling application will be managing - in such cases it's better to provide slower behaviour that never gets too bad... so lean towards shrinking the memory blocks to fit the string data (a set number of additional operations so doesn't affect big-O efficiency) rather than having an unknown proportion of the original string buffer wasted (in a pathological case - zero or one character used after arbitrarily large allocations). As a performance optimisation you might only bother returning memory if unusued space is >= the used space - tune to taste, or make it caller-configurable.
You comment on another answer:
So it comes down to judging whether the realloc will take longer, or the preprocessing size determination?
If performance is your top priority, then yes - you'd want to profile. If you're not CPU bound, then as a general rule take the "preprocessing" hit and do a right-sized allocation - there's just less fragmentation and mess. Countering that, if you have to write a special preprocessing mode for some function - that's an extra "surface" for errors and code to maintain. (This trade-off decision is commonly needed when implementing your own asprintf() from snprintf(), but there at least you can trust snprintf() to act as documented and don't personally have to maintain it).
Once '\0' is added, does the memory just get returned, or is it
sitting there hogging space until free() is called?
There's nothing magical about \0. You have to call realloc if you want to "shrink" the allocated memory. Otherwise the memory will just sit there until you call free.
If I stick a '\0' into the middle of the allocated memory, does (a.)
free() still work properly
Whatever you do in that memory free will always work properly if you pass it the exact same pointer returned by malloc. Of course if you write outside it all bets are off.
\0 is just one more character from malloc and free perspective, they don't care what data you put in the memory. So free will still work whether you add \0 in the middle or don't add \0 at all. The extra space allocated will still be there, it won't be returned back to the process as soon as you add \0 to the memory. I personally would prefer to allocate only the required amount of memory instead of allocating at some upper bound as that will just wasting the resource.
As soon as you get memory from heap by calling malloc(), the memory is yours to use. Inserting \0 is like inserting any other character. This memory will remain in your possession until you free it or until OS claims it back.
The \0is a pure convention to interpret character arrays as stings - it is independent of the memory management. I.e., if you want to get your money back, you should call realloc. The string does not care about memory (what is a source of many security problems).
malloc just allocates a chunk of memory .. Its upto you to use however you want and call free from the initial pointer position... Inserting '\0' in the middle has no consequence...
To be specific malloc doesnt know what type of memory you want (It returns onle a void pointer) ..
Let us assume you wish to allocate 10 bytes of memory starting 0x10 to 0x19 ..
char * ptr = (char *)malloc(sizeof(char) * 10);
Inserting a null at 5th position (0x14) does not free the memory 0x15 onwards...
However a free from 0x10 frees the entire chunk of 10 bytes..
free() will still work with a NUL byte in memory
the space will remain wasted until free() is called, or unless you subsequently shrink the allocation
Generally, memory is memory is memory. It doesn't care what you write into it. BUT it has a race, or if you prefer a flavor (malloc, new, VirtualAlloc, HeapAlloc, etc). This means that the party that allocates a piece of memory must also provide the means to deallocate it. If your API comes in a DLL, then it should provide a free function of some sort.
This of course puts a burden on the caller right?
So why not put the WHOLE burden on the caller?
The BEST way to deal with dynamically allocated memory is to NOT allocate it yourself. Have the caller allocate it and pass it on to you. He knows what flavor he allocated, and he is responsible to free it whenever he is done using it.
How does the caller know how much to allocate?
Like many Windows APIs have your function return the required amount of bytes when called e.g. with a NULL pointer, then do the job when provided with a non-NULL pointer (using IsBadWritePtr if it is suitable for your case to double-check accessibility).
This can also be much much more efficient. Memory allocations COST a lot. Too many memory allocations cause heap fragmentation and then the allocations cost even more. That's why in kernel mode we use the so called "look-aside lists". To minimize the number of memory allocations done, we reuse the blocks we have already allocated and "freed", using services that the NT Kernel provides to driver writers.
If you pass on the responsibility for memory allocation to your caller, then he might be passing you cheap memory from the stack (_alloca), or passing you the same memory over and over again without any additional allocations. You don't care of course, but you DO allow your caller to be in charge of optimal memory handling.
To elaborate on the use of the NULL terminator in C:
You cannot allocate a "C string" you can allocate a char array and store a string in it, but malloc and free just see it as an array of the requested length.
A C string is not a data type but a convention for using a char array where the null character '\0' is treated as the string terminator.
This is a way to pass strings around without having to pass a length value as a separate argument. Some other programming languages have explicit string types that store a length along with the character data to allow passing strings in a single parameter.
Functions that document their arguments as "C strings" are passed char arrays but have no way of knowing how big the array is without the null terminator so if it is not there things will go horribly wrong.
You will notice functions that expect char arrays that are not necessarily treated as strings will always require a buffer length parameter to be passed.
For example if you want to process char data where a zero byte is a valid value you can't use '\0' as a terminator character.
You could do what some of the MS Windows APIs do where you (the caller) pass a pointer and the size of the memory you allocated. If the size isn't enough, you're told how many bytes to allocate. If it was enough, the memory is used and the result is the number of bytes used.
Thus the decision about how to efficiently use memory is left to the caller. They can allocate a fixed 255 bytes (common when working with paths in Windows) and use the result from the function call to know whether more bytes are needed (not the case with paths due to MAX_PATH being 255 without bypassing Win32 API) or whether most of the bytes can be ignored...
The caller could also pass zero as the memory size and be told exactly how much needs to be allocated - not as efficient processing-wise, but could be more efficient space-wise.
You can certainly preallocate to an upperbound, and use all or something less.
Just make sure you actually use all or something less.
Making two passes is also fine.
You asked the right questions about the tradeoffs.
How do you decide?
Use two passes, initially, because:
1. you'll know you aren't wasting memory.
2. you're going to profile to find out where
you need to optimize for speed anyway.
3. upperbounds are hard to get right before
you've written and tested and modified and
used and updated the code in response to new
requirements for a while.
4. simplest thing that could possibly work.
You might tighten up the code a little, too.
Shorter is usually better. And the more the
code takes advantage of known truths, the more
comfortable I am that it does what it says.
char* copyWithoutDuplicateChains(const char* str)
{
if (str == NULL) return NULL;
const char* s = str;
char prev = *s; // [prev][s+1]...
unsigned int outlen = 1; // first character counted
// Determine length necessary by mimicking processing
while (*s)
{ while (*++s == prev); // skip duplicates
++outlen; // new character encountered
prev = *s; // restart chain
}
// Construct output
char* outstr = (char*)malloc(outlen);
s = str;
*outstr++ = *s; // first character copied
while (*s)
{ while (*++s == prev); // skip duplicates
*outstr++ = *s; // copy new character
}
// done
return outstr;
}
I basically have this piece of code.
char (* text)[1][80];
text = calloc(2821522,80);
The way I calculated it, that calloc should have allocated 215.265045 megabytes of RAM, however, the program in the end exceeded that number and allocated nearly 700mb of ram.
So it appears I cannot properly know how much memory that function will allocate.
How does one calculate that propery?
calloc (and malloc for that matter) is free to allocate as much space as it needs to satisfy the request.
So, no, you cannot tell in advance how much it will actually give you, you can only assume that it's given you the amount you asked for.
Having said that, 700M seems a little excessive so I'd be investigating whether the calloc was solely responsible for that by, for example, a program that only does the calloc and nothing more.
You might also want to investigate how you're measuring that memory usage.
For example, the following program:
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
int main (void) {
char (* text)[1][80];
struct mallinfo mi;
mi = mallinfo(); printf ("%d\n", mi.uordblks);
text = calloc(2821522,80);
mi = mallinfo(); printf ("%d\n", mi.uordblks);
return 0;
}
outputs, on my system:
66144
225903256
meaning that the calloc has allocated 225,837,112 bytes which is only a smidgeon (115,352 bytes or 0.05%) above the requested 225,721,760.
Well it depends on the underlying implementation of malloc/calloc.
It generally works like this - there's this thing called the heap pointer which points to the top of the heap - the area from where dynamic memory gets allocated. When memory is first allocated, malloc internally requests x amount of memory from the kernel - i.e. the heap pointer increments by a certain amount to make that space available. That x may or may not be equal to the size of the memory block you requested (it might be larger to account for future mallocs). If it isn't, then you're given at least the amount of memory you requested(sometimes you're given more memory because of alignment issues). The rest is made part of an internal free list maintained by malloc. To sum it up malloc has some underlying data structures and a lot depends on how they are implemented.
My guess is that the x amount of memory was larger (for whatever reason) than you requested and hence malloc/calloc was holding on to the rest in its free list. Try allocating some more memory and see if the footprint increases.
For a small to-be-embedded application, I wrote a few functions + struct that work as String Buffer (similar to std::stringstream in C++).
While the code as such works fine, There are a few not-so-minor problems:
I never before wrote functions in C that manually allocate and use growing memory, thus I'm afraid there are still some quirks that yet need to be adressed
It seems the code allocates far more memory than it actually needs, which is VERY BAD
Due to warnings reported by valgrind I have switched from malloc to calloc in one place in the code, which sucessfully removed the warning, but I'm not entirely sure if i'm actually using it correctly
Example of what I mean that it allocates more than it really needs (using a 56k file):
==23668== HEAP SUMMARY:
==23668== in use at exit: 0 bytes in 0 blocks
==23668== total heap usage: 49,998 allocs, 49,998 frees, 1,249,875,362 bytes allocated
... It just doesn't look right ...
The code in question is here (too large to copy it in a <code> field on SO): http://codepad.org/LQzphUzd
Help is needed, and I'm grateful for any advice!
The way you are growing your buffer is rather inefficient. For each little piece of string, you realloc() memory, which can mean new memory is allocated and the contents of the "old" memory are copied over. That is slow and fragments your heap.
Better is to grow in fixed amounts, or in fixed percentages, i.e. make the new size 1.5 or 2 times the size of the old size. That also wastes some memory, but will keep the heap more usable and not so many copies are made.
This means you'll have to keep track of two values: capacity (number of bytes allocated) and length (actual length of the string). But that should not be too hard.
I would introduce a function "FstrBuf_Grow" which takes care of all of that. You just call it with the amount of memory you want to add, and FstrBuf_Grow will take care that the capacity matches the requirements by reallocing when necessary and at least as much as necessary.
...
void FstrBuf_Grow(FstringBuf *buf, size_t more)
{
while (buf->length + more) > buf->capacity
buf->capacity = 3 * buf->capacity / 2;
buf->data = realloc(buf->data, buf->capacity + 1);
}
That multiplies capacity by 1.5 until data is large enough. You can choose different strategies, depending on your needs.
The strncat(ptr->data, str, len);, move before the ptr->length = ((ptr->length) + len); and use strncpy(ptr->data+ptr->length.... And the ptr = NULL; in the Destroy is useless.
The code of the "library" seems to be correct BUT be aware that you are continously reallocating the buffer. Normally you should try to grow the buffer only rarely (for example every time you need to grow the buffer you use max(2* the current size, 4) as the new size) because growing the buffer is O(n). The big memory allocation is probably because the first time you allocate a small buffer. Then you realloc it in a bigger buffer. Then you need to realloc it in a buffer even bigger and so the heap grows.
It looks like you're re-allocating the buffer on every append. Shouldn't you grow it only when you want to append more than it can hold?
When reallocating you want to increase the size of the buffer using a strategy that gives you the best trade off between the number of allocations and the amount of memory allocated. Just doubling the size of the buffer every time you hit the limit might not be ideal for an embedded program.
Generally for embedded applications it is much better to allocate a circular FIFO buffer 1-3 times the maximum message size.
int main()
{
char *s1, *sTemp;
s1 = (char*)malloc(sizeof(char)*7);
*(s1 + 0) = 'a';
*(s1 + 1) = 'b';
*(s1 + 2) = 'c';
*(s1 + 3) = 'd';
*(s1 + 4) = 'e';
*(s1 + 5) = 'f';
*(s1 + 6) = '\0';
sTemp = (s1 + 3);
free(sTemp); // shud delete d onwards. But it doesn't !!
return 0;
}
Hello,
In the C above code sTemp should point to the 3rd cell beyond s1 ( occupied by 'd')
So on calling free(sTemp) i expect to have something deleted from this location onwards.
( I purposely mention 'something' as the motive of my experiment initially was to find out till which location the free() ing works )
However i recieve a SIGABRT at the free().
How does free() know that this is not the start of the chunk. and correspondingly can we free up memory only from start of chunks? [ are they only the free-able pointers that free() can accept?? ]
Looking forward to replies :)
From the man pages
free() frees the memory space pointed
to by ptr, which must have been
returned by a previous call to
malloc(), calloc() or realloc().
Otherwise, or if free(ptr) has already
been called before, undefined
behaviour occurs.
Source: http://linux.die.net/man/3/free
About the actual question "how does free know...":
free knows that it is not at the start of the chunk because it maintains metadata that you don't know about. If you think about it, that's necessarily so, because otherwise it could never know how much memory to free, given only an address.
It is not specified how exactly the runtime keeps book of allocation sizes, only that you must never pass any pointer to free that did not come from a function of the malloc family.
Usually this works by malloc allocating a little more memory than needed[1] and writing some metadata in memory preceding the address that is returned. free then just looks at e.g. ptr[-8] or whatever (implementation detail!) to read this metadata.
It can then optionally do some consistency checks, and can determine the size of the memory block (one trivial consistency check that is probably always done is checking proper alignment of the pointer).
Having mentioned that, please please please, don't even think about playing with this metadata.
[1] It often does that anyway to satisfy alignment requirements and because of allocator internals (most allocators manage memory in different "buckets" of fixed size), so if you allocate, say, 4 bytes, you nominally get 4 bytes, but the runtime really allocated 8 or 16 bytes most of the time (without you knowing).
You can only free() pointers that were actually malloced (or calloced, realloced). If you try to free only a portion of memory by passing in a different pointer (even one that is "part" of another pointer), the C runtime will not be pleased (as you can see.)
I think you cannot free the memory because free doesn't know how much memory to free.Your program has information about the address malloc() returned and not for every address in this space.So you can free(s1) but not free(s1+3).Also you can handle your pointers as an array in this example:
s1[0]='a';
You can only free() memory that was previously malloced, calloced, or realloced, as dlev and Daniel have said.
The reason for this is implementation-dependent, but involves the method of keeping track of allocated memory.
Efficient memory allocation is a difficult problem because different memory allocation algorithms work well depending upon how memory is allocated: a few small chunks, half of which are freed, then slightly larger blocks being grabbed, etc.
The objective is to keep the size of the memory block used by the program at a minimum (usually this chunk will be a continuous block of virtual memory), while keeping the usage of the space within that block extremely high (few gaps between used segments of memory).
Remember that these blocks can't be moved except when being realloced, so there's always going to be some wasted space.
To minimize the waste, metadata about (at least) the size of the block is stored just before it. The memory allocator can look through the used blocks when determining how to handle a new request. If you pick a random memory location, whether part of a previously-allocated region or no, this metadata will not be present and free will be unable to determine what should be freed.
You cannot free that pointer the way you are doing, check this question and its answer: Memory Clobbering Error