what's the point in malloc(0)?

问题:

Just saw this code:

artist = (char *) malloc(0);

and I was wondering why would one do this?

回答1:

According to the specifications, malloc(0) will return either \"a null pointer or a unique pointer that can be successfully passed to free()\".

This basically lets you allocate nothing, but still pass the \"artist\" variable to a call to free() without worry. For practical purposes, it\'s pretty much the same as doing:

artist = NULL;

回答2:

The C standard says:

If the space cannot be allocated, a null pointer is returned. If the size of the space requested is zero, the behavior is implementation defined: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.

So, malloc(0) could return NULL or a valid pointer that may not be dereferenced. In either case, it\'s perfectly valid to call free() on it.

I don\'t really think malloc(0) has much use, except in cases when malloc(n) is called in a loop for example, and n might be zero.

Looking at the code in the link, I believe that the author had two misconceptions:

• malloc(0) returns a valid pointer always, and
• free(0) is bad.

So, he made sure that artist and other variables always had some \"valid\" value in them. The comment says as much: // these must always point at malloc\'d data.

回答3:

malloc(0) behaviour is implementation specific. The library can return NULL or have the regular malloc behaviour, with no memory allocated. Whatever it does, it must be documented somewhere.

Usually, it returns a pointer that is valid and unique but should NOT be dereferenced. Also note that it CAN consume memory even though it did not actually allocate anything.

It is possible to realloc a non null malloc(0) pointer.

Having a malloc(0) verbatim is not much use though. It\'s mostly used when a dynamic allocation is zero byte and you didn\'t care to validate it.

回答4:

There\'s an answer elsewhere on this page that begins \"malloc(0) will return a valid memory address and whose range will depend on the type of pointer which is being allocated memory\". This statement is incorrect (I don\'t have enough reputation to comment on that answer directly, so can\'t put this comment directly under there).

Doing malloc(0) will not automatically allocate memory of correct size. The malloc function is unaware of what you\'re casting its result to. The malloc function relies purely on the size number that you give as its argument. You need to do malloc(sizeof(int)) to get enough storage to hold an int, for example, not 0.

回答5:

There are a lot of half true answers around here, so here are the hard facts. The man-page for malloc() says:

If size is 0, then malloc() returns either NULL, or a unique pointer value that can later be successfully passed to free().

That means, there is absolutely no guarantee that the result of malloc(0) is either unique or not NULL. The only guarantee is provided by the definition of free(), again, here is what the man-page says:

If ptr is NULL, no operation is performed.

So, whatever malloc(0) returns, it can safely be passed to free(). But so can a NULL pointer.

Consequently, writing artist = malloc(0); is in no way better than writing artist = NULL;

回答6:

malloc(0) doesn\'t make any sense to me, unless the code is relying on behaviour specific to the implementation. If the code is meant to be portable, then it has to account for the fact that a NULL return from malloc(0) isn\'t a failure. So why not just assign NULL to artist anyway, since that\'s a valid successful result, and is less code, and won\'t cause your maintenance programmers to take time figuring it out?

malloc(SOME_CONSTANT_THAT_MIGHT_BE_ZERO) or malloc(some_variable_which_might_be_zero) perhaps could have their uses, although again you have to take extra care not to treat a NULL return as a failure if the value is 0, but a 0 size is supposed to be OK.

回答7:

Admittedly, I have never seen this before, this is the first time I\'ve seen this syntax, one could say, a classic case of function overkill. In conjunction to Reed\'s answer, I would like to point out that there is a similar thing, that appears like an overloaded function realloc:

• foo is non-NULL and size is zero, realloc(foo, size);. When you pass in a non-NULL pointer and size of zero to realloc, realloc behaves as if you’ve called free(…)
• foo is NULL and size is non-zero and greater than 1, realloc(foo, size);. When you pass in a NULL pointer and size is non-zero, realloc behaves as if you’ve called malloc(…)

Hope this helps, Best regards, Tom.

回答8:

To actually answer the question made: there is no reason to do that

回答9:

Why you shouldn\'t do this...

Since malloc\'s return value is implementation dependent, you may get a NULL pointer or some other address back. This can end up creating heap-buffer overflows if error handling code doesn\'t check both size and returned value, leading to stability issues (crashes) or even worse security issues.

Consider this example, where further accessing memory via returned address will corrupt heap iff size is zero and implementation returns a non NULL value back.

size_t size;

/* Initialize size, possibly by user-controlled input */

int *list = (int *)malloc(size);
if (list == NULL) {
  /* Handle allocation error */
}
else {
  /* Continue processing list */
}

See this Secure Coding page from CERT Coding Standards where I took the example above for further reading.

回答10:

Not sure, according to some random malloc source code I found, an input of 0 results in a return value of NULL. So it\'s a crazy way of setting the artist pointer to NULL.

http://www.raspberryginger.com/jbailey/minix/html/lib_2ansi_2malloc_8c-source.html

回答11:

malloc(0) will return NULL or a valid pointer which can be rightly passed to free. And though it seems like the memory that it points to is useless or it can\'t be written to or read from, that is not always true. :)

int *i = malloc(0);
*i = 100;
printf(\"%d\", *i);

We expect a segmentation fault here, but surprisingly, this prints 100! It is because malloc actually asks for a huge chunk of memory when we call malloc for the first time. Every call to malloc after that, uses memory from that big chunk. Only after that huge chunk is over, new memory is asked for.

Use of malloc(0): if you are in a situation where you want subsequent malloc calls to be faster, calling malloc(0) should do it for you (except for edge cases).

回答12:

In Windows:

• void *p = malloc(0); will allocate a zero-length buffer on the local heap. The pointer returned is a valid heap pointer.
• malloc ultimately calls HeapAlloc using the default C runtime heap which then calls RtlAllocateHeap, etc.
• free(p); uses HeapFree to free the 0-length buffer on the heap. Not freeing it would result in a memory leak.

回答13:

Here is the analysis after running with valgrind memory check tool.

==16740== Command: ./malloc0
==16740==
p1 = 0x5204040
==16740==
==16740== HEAP SUMMARY:
==16740==     in use at exit: 0 bytes in 0 blocks
==16740==   total heap usage: 2 allocs, 2 frees, 1,024 bytes allocated
==16740==
==16740== All heap blocks were freed -- no leaks are possible

and here\'s my sample code:

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>

int main()
{
   //int i;
   char *p1;

   p1 = (char *)malloc(0);
   printf(\"p1 = %p\\n\", p1);

   free(p1);

   return 0;

}

By default 1024 bytes is allocated. If I increase the size of malloc, the allocated bytes will increase by 1025 and so on.

回答14:

According to Reed Copsey answer and the man page of malloc , I wrote some examples to test. And I found out malloc(0) will always give it a unique value. See my example :

char *ptr;
if( (ptr = (char *) malloc(0)) == NULL )
    puts(\"Got a null pointer\");
else
    puts(\"Got a valid pointer\");

The output will be \"Got a valid pointer\", which means ptr is not null.

回答15:

malloc(0) will return a valid memory address and whose range will depend on the type of pointer which is being allocated memory. Also you can assign values to the memory area but this should be in range with the type of pointer being used. You can also free the allocated memory. I will explain this with an example:

int *p=NULL;
p=(int *)malloc(0);
free(p);

The above code will work fine in a gcc compiler on Linux machine. If you have a 32 bit compiler then you can provide values in the integer range, i.e. -2147483648 to 2147483647. Same applies for characters also. Please note that if type of pointer declared is changed then range of values will change regardless of malloc typecast, i.e.

unsigned char *p=NULL;
p =(char *)malloc(0);
free(p);

p will take a value from 0 to 255 of char since it is declared an unsigned int.

回答16:

Just to correct a false impression here:

artist = (char *) malloc(0); will never ever return NULL; it\'s not the same as artist = NULL;. Write a simple program and compare artist with NULL. if (artist == NULL) is false and if (artist) is true.