Difference between malloc and calloc?

Question

What is the difference between doing:

ptr = malloc(MAXELEMS * sizeof(char *));

And:

ptr = calloc(MAXELEMS, sizeof(char*));

When is it a good idea to use calloc over malloc or vice versa?

Answer 1

calloc() gives you a zero-initialized buffer, while malloc() leaves the memory uninitialized.

For large allocations, most calloc implementations under mainstream OSes will get known-zeroed pages from the OS (e.g. via POSIX mmap(MAP_ANONYMOUS) or Windows VirtualAlloc) so it doesn't need to write them in user-space. This is how normal malloc gets more pages from the OS as well; calloc just takes advantage of the OS's guarantee.

This means calloc memory can still be "clean" and lazily-allocated, and copy-on-write mapped to a system-wide shared physical page of zeros. (Assuming a system with virtual memory.) The effects are visible with performance experiments on Linux, for example.

Some compilers even can optimize malloc + memset(0) into calloc for you, but it's best to just use calloc in the source if you want zeroed memory. (Or if you were trying to pre-fault it to avoid page faults later, that optimization will defeat your attempt.)

If you aren't going to ever read memory before writing it, use malloc so it can (potentially) give you dirty memory from its internal free list instead of getting new pages from the OS. (Or instead of zeroing a block of memory on the free list for a small allocation).

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Embedded implementations of calloc may leave it up to calloc itself to zero memory if there's no OS, or it's not a fancy multi-user OS that zeros pages to stop information leaks between processes.

On embedded Linux, malloc could mmap(MAP_UNINITIALIZED|MAP_ANONYMOUS), which is only enabled for some embedded kernels because it's insecure on a multi-user system.

Answer 2

A less known difference is that in operating systems with optimistic memory allocation, like Linux, the pointer returned by malloc isn't backed by real memory until the program actually touches it.

calloc does indeed touch the memory (it writes zeroes on it) and thus you'll be sure the OS is backing the allocation with actual RAM (or swap). This is also why it is slower than malloc (not only does it have to zero it, the OS must also find a suitable memory area by possibly swapping out other processes)

See for instance this SO question for further discussion about the behavior of malloc

Answer 3

One often-overlooked advantage of calloc is that (conformant implementations of) it will help protect you against integer overflow vulnerabilities. Compare:

size_t count = get_int32(file);
struct foo *bar = malloc(count * sizeof *bar);

vs.

size_t count = get_int32(file);
struct foo *bar = calloc(count, sizeof *bar);

The former could result in a tiny allocation and subsequent buffer overflows, if count is greater than SIZE_MAX/sizeof *bar. The latter will automatically fail in this case since an object that large cannot be created.

Of course you may have to be on the lookout for non-conformant implementations which simply ignore the possibility of overflow... If this is a concern on platforms you target, you'll have to do a manual test for overflow anyway.

Answer 4

The documentation makes the calloc look like malloc, which just does zero-initialize the memory; this is not the primary difference! The idea of calloc is to abstract copy-on-write semantics for memory allocation. When you allocate memory with calloc it all maps to same physical page which is initialized to zero. When any of the pages of the allocated memory is written into a physical page is allocated. This is often used to make HUGE hash tables, for example since the parts of hash which are empty aren't backed by any extra memory (pages); they happily point to the single zero-initialized page, which can be even shared between processes.

Any write to virtual address is mapped to a page, if that page is the zero-page, another physical page is allocated, the zero page is copied there and the control flow is returned to the client process. This works same way memory mapped files, virtual memory, etc. work.. it uses paging.

Here is one optimization story about the topic: http://blogs.fau.de/hager/2007/05/08/benchmarking-fun-with-calloc-and-zero-pages/

Answer 5

There's no difference in the size of the memory block allocated. calloc just fills the memory block with physical all-zero-bits pattern. In practice it is often assumed that the objects located in the memory block allocated with calloc have initilial value as if they were initialized with literal 0, i.e. integers should have value of 0, floating-point variables - value of 0.0, pointers - the appropriate null-pointer value, and so on.

From the pedantic point of view though, calloc (as well as memset(..., 0, ...)) is only guaranteed to properly initialize (with zeroes) objects of type unsigned char. Everything else is not guaranteed to be properly initialized and may contain so called trap representation, which causes undefined behavior. In other words, for any type other than unsigned char the aforementioned all-zero-bits patterm might represent an illegal value, trap representation.

Later, in one of the Technical Corrigenda to C99 standard, the behavior was defined for all integer types (which makes sense). I.e. formally, in the current C language you can initialize only integer types with calloc (and memset(..., 0, ...)). Using it to initialize anything else in general case leads to undefined behavior, from the point of view of C language.

In practice, calloc works, as we all know :), but whether you'd want to use it (considering the above) is up to you. I personally prefer to avoid it completely, use malloc instead and perform my own initialization.

Finally, another important detail is that calloc is required to calculate the final block size internally, by multiplying element size by number of elements. While doing that, calloc must watch for possible arithmetic overflow. It will result in unsuccessful allocation (null pointer) if the requested block size cannot be correctly calculated. Meanwhile, your malloc version makes no attempt to watch for overflow. It will allocate some "unpredictable" amount of memory in case overflow happens.

Answer 6

from an article Benchmarking fun with calloc() and zero pages on Georg Hager's Blog

When allocating memory using calloc(), the amount of memory requested is not allocated right away. Instead, all pages that belong to the memory block are connected to a single page containing all zeroes by some MMU magic (links below). If such pages are only read (which was true for arrays b, c and d in the original version of the benchmark), the data is provided from the single zero page, which – of course – fits into cache. So much for memory-bound loop kernels. If a page gets written to (no matter how), a fault occurs, the “real” page is mapped and the zero page is copied to memory. This is called copy-on-write, a well-known optimization approach (that I even have taught multiple times in my C++ lectures). After that, the zero-read trick does not work any more for that page and this is why performance was so much lower after inserting the – supposedly redundant – init loop.

Answer 7

Number of blocks:
malloc() assigns single block of requested memory,
calloc() assigns multiple blocks of the requested memory

Initialization:
malloc() - doesn't clear and initialize the allocated memory.
calloc() - initializes the allocated memory by zero.

Speed:
malloc() is fast.
calloc() is slower than malloc().

Arguments & Syntax:
malloc() takes 1 argument:

1. bytes

   • The number of bytes to be allocated

calloc() takes 2 arguments:

1. length

   • the number of blocks of memory to be allocated

2. bytes

   • the number of bytes to be allocated at each block of memory

void *malloc(size_t bytes);         
void *calloc(size_t length, size_t bytes);      

Manner of memory Allocation:
The malloc function assigns memory of the desired 'size' from the available heap.
The calloc function assigns memory that is the size of what’s equal to ‘num *size’.

Meaning on name:
The name malloc means "memory allocation".
The name calloc means "contiguous allocation".

Answer 8

calloc is generally malloc+memset to 0

It is generally slightly better to use malloc+memset explicitly, especially when you are doing something like:

ptr=malloc(sizeof(Item));
memset(ptr, 0, sizeof(Item));

That is better because sizeof(Item) is know to the compiler at compile time and the compiler will in most cases replace it with the best possible instructions to zero memory. On the other hand if memset is happening in calloc, the parameter size of the allocation is not compiled in in the calloc code and real memset is often called, which would typically contain code to do byte-by-byte fill up until long boundary, than cycle to fill up memory in sizeof(long) chunks and finally byte-by-byte fill up of the remaining space. Even if the allocator is smart enough to call some aligned_memset it will still be a generic loop.

One notable exception would be when you are doing malloc/calloc of a very large chunk of memory (some power_of_two kilobytes) in which case allocation may be done directly from kernel. As OS kernels will typically zero out all memory they give away for security reasons, smart enough calloc might just return it withoud additional zeroing. Again - if you are just allocating something you know is small, you may be better off with malloc+memset performance-wise.

Answer 9

There are two differences.
First, is in the number of arguments. malloc() takes a single argument (memory required in bytes), while calloc() needs two arguments.
Secondly, malloc() does not initialize the memory allocated, while calloc() initializes the allocated memory to ZERO.

• calloc() allocates a memory area, the length will be the product of its parameters. calloc fills the memory with ZERO's and returns a pointer to first byte. If it fails to locate enough space it returns a NULL pointer.

Syntax: ptr_var = calloc(no_of_blocks, size_of_each_block); i.e. ptr_var = calloc(n, s);

• malloc() allocates a single block of memory of REQUSTED SIZE and returns a pointer to first byte. If it fails to locate requsted amount of memory it returns a null pointer.

Syntax: ptr_var = malloc(Size_in_bytes); The malloc() function take one argument, which is the number of bytes to allocate, while the calloc() function takes two arguments, one being the number of elements, and the other being the number of bytes to allocate for each of those elements. Also, calloc() initializes the allocated space to zeroes, while malloc() does not.

Answer 10

Difference 1:

malloc() usually allocates the memory block and it is initialized memory segment.

calloc() allocates the memory block and initialize all the memory block to 0.

Difference 2:

If you consider malloc() syntax, it will take only 1 argument. Consider the following example below:

data_type ptr = (cast_type *)malloc( sizeof(data_type)*no_of_blocks );

Ex: If you want to allocate 10 block of memory for int type,

int *ptr = (int *) malloc(sizeof(int) * 10 );

If you consider calloc() syntax, it will take 2 arguments. Consider the following example below:

data_type ptr = (cast_type *)calloc(no_of_blocks, (sizeof(data_type)));

Ex: if you want to allocate 10 blocks of memory for int type and Initialize all that to ZERO,

int *ptr = (int *) calloc(10, (sizeof(int)));

Similarity:

Both malloc() and calloc() will return void* by default if they are not type casted .!

Answer 11

The calloc() function that is declared in the <stdlib.h> header offers a couple of advantages over the malloc() function.

1. It allocates memory as a number of elements of a given size, and
2. It initializes the memory that is allocated so that all bits are zero.

Answer 12

malloc() and calloc() are functions from the C standard library that allow dynamic memory allocation, meaning that they both allow memory allocation during runtime.

Their prototypes are as follows:

void *malloc( size_t n);
void *calloc( size_t n, size_t t)

There are mainly two differences between the two:

• Behavior: malloc() allocates a memory block, without initializing it, and reading the contents from this block will result in garbage values. calloc(), on the other hand, allocates a memory block and initializes it to zeros, and obviously reading the content of this block will result in zeros.

• Syntax: malloc() takes 1 argument (the size to be allocated), and calloc() takes two arguments (number of blocks to be allocated and size of each block).

The return value from both is a pointer to the allocated block of memory, if successful. Otherwise, NULL will be returned indicating the memory allocation failure.

Example:

int *arr;

// allocate memory for 10 integers with garbage values
arr = (int *)malloc(10 * sizeof(int)); 

// allocate memory for 10 integers and sets all of them to 0
arr = (int *)calloc(10, sizeof(int));

The same functionality as calloc() can be achieved using malloc() and memset():

// allocate memory for 10 integers with garbage values   
arr= (int *)malloc(10 * sizeof(int));
// set all of them to 0
memset(arr, 0, 10 * sizeof(int)); 

Note that malloc() is preferably used over calloc() since it's faster. If zero-initializing the values is wanted, use calloc() instead.

Answer 13

A difference not yet mentioned: size limit

void *malloc(size_t size) can only allocate up to SIZE_MAX.

void *calloc(size_t nmemb, size_t size); can allocate up about SIZE_MAX*SIZE_MAX.

This ability is not often used in many platforms with linear addressing. Such systems limit calloc() with nmemb * size <= SIZE_MAX.

Consider a type of 512 bytes called disk_sector and code wants to use lots of sectors. Here, code can only use up to SIZE_MAX/sizeof disk_sector sectors.

size_t count = SIZE_MAX/sizeof disk_sector;
disk_sector *p = malloc(count * sizeof *p);

Consider the following which allows an even larger allocation.

size_t count = something_in_the_range(SIZE_MAX/sizeof disk_sector + 1, SIZE_MAX)
disk_sector *p = calloc(count, sizeof *p);

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Now if such a system can supply such a large allocation is another matter. Most today will not. Yet it has occurred for many years when SIZE_MAX was 65535. Given Moore's law, suspect this will be occurring about 2030 with certain memory models with SIZE_MAX == 4294967295 and memory pools in the 100 of GBytes.

Answer 14

Both malloc and calloc allocate memory, but calloc initialises all the bits to zero whereas malloc doesn't.

Calloc could be said to be equivalent to malloc + memset with 0 (where memset sets the specified bits of memory to zero).

So if initialization to zero is not necessary, then using malloc could be faster.