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Full Discussion: Malloc to void pointer fails
Top Forums Programming Malloc to void pointer fails Post 302910794 by jim mcnamara on Monday 28th of July 2014 09:12:41 AM
Old 07-28-2014
There is a lot of opinion in this thread, which is okay as long as a reader knows that fact.

Generally it is bad idea to cast malloc, because it is not required and may introduce subtle bugs that are hard to find, a discussion:

FAQ > Casting malloc - Cprogramming.com

Consider the use of a debugger, ex: gdb. This will help resolve crashes by examining core files, link:

RMS's gdb Tutorial: Segmentation Fault Example

Finally, allocating heap memory (malloc does this for you in C) is not all that simple.
If you allocate x+1 bytes for a string which should never be long than x, you have to check input carefully, otherwise if the string you enter is too long by a few bytes it probably will not segfault, it will simply trash a neighboring variable. Nasty.

Except for embedded systems (ex ARM), declaring strings longer than needed is less harmful, but still requires checking EVERY input string before parking it in the variable, because this makes an entry for a possible code exploit. Shell code and SQL injection come to mind.

malloc works this way in general:
1. at the beginning of code invocation, the brk() system call allocates pages of memory, and those pages then are controlled by malloc, not directly by your program code, normally. Do not call brk() on your own if you use malloc or functions like strdup which call malloc. Calling brk() directly in code that uses malloc usually results in chaos.

So if you malloc 10 bytes and page size is 8192 bytes (example), then you still have more already allocated memory available.

2. malloc keeps track of what it allocates, pages are in a page table, variables are tracked usually with some kind of descriptor. The descriptor is often a struct that consists of a pointer to the start of the variable (what malloc returns to you), and a length in bytes.

hypothetical example base on Doug Lea's original malloc:

[address of word aligned memory][length].... [word aligned memory]

So, if you increment or decrement the pointer you have (your variable) it no longer references [address of word aligned memory]. This causes free() to get nasty with you.

- finally, some of the suggestions in this thread are just that - there is both art and science in building good code.
 

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MALLOC(3)						     Linux Programmer's Manual							 MALLOC(3)

NAME
calloc, malloc, free, realloc - Allocate and free dynamic memory SYNOPSIS
#include <stdlib.h> void *calloc(size_t nmemb, size_t size); void *malloc(size_t size); void free(void *ptr); void *realloc(void *ptr, size_t size); DESCRIPTION
calloc() allocates memory for an array of nmemb elements of size bytes each and returns a pointer to the allocated memory. The memory is set to zero. malloc() allocates size bytes and returns a pointer to the allocated memory. The memory is not cleared. free() frees the memory space pointed to by ptr, which must have been returned by a previous call to malloc(), calloc() or realloc(). Oth- erwise, or if free(ptr) has already been called before, undefined behaviour occurs. If ptr is NULL, no operation is performed. realloc() changes the size of the memory block pointed to by ptr to size bytes. The contents will be unchanged to the minimum of the old and new sizes; newly allocated memory will be uninitialized. If ptr is NULL, the call is equivalent to malloc(size); if size is equal to zero, the call is equivalent to free(ptr). Unless ptr is NULL, it must have been returned by an earlier call to malloc(), calloc() or realloc(). RETURN VALUE
For calloc() and malloc(), the value returned is a pointer to the allocated memory, which is suitably aligned for any kind of variable, or NULL if the request fails. free() returns no value. realloc() returns a pointer to the newly allocated memory, which is suitably aligned for any kind of variable and may be different from ptr, or NULL if the request fails. If size was equal to 0, either NULL or a pointer suitable to be passed to free() is returned. If real- loc() fails the original block is left untouched - it is not freed or moved. CONFORMING TO
ANSI-C SEE ALSO
brk(2), posix_memalign(3) NOTES
The Unix98 standard requires malloc(), calloc(), and realloc() to set errno to ENOMEM upon failure. Glibc assumes that this is done (and the glibc versions of these routines do this); if you use a private malloc implementation that does not set errno, then certain library routines may fail without having a reason in errno. Crashes in malloc(), free() or realloc() are almost always related to heap corruption, such as overflowing an allocated chunk or freeing the same pointer twice. Recent versions of Linux libc (later than 5.4.23) and GNU libc (2.x) include a malloc implementation which is tunable via environment vari- ables. When MALLOC_CHECK_ is set, a special (less efficient) implementation is used which is designed to be tolerant against simple errors, such as double calls of free() with the same argument, or overruns of a single byte (off-by-one bugs). Not all such errors can be protected against, however, and memory leaks can result. If MALLOC_CHECK_ is set to 0, any detected heap corruption is silently ignored; if set to 1, a diagnostic is printed on stderr; if set to 2, abort() is called immediately. This can be useful because otherwise a crash may happen much later, and the true cause for the problem is then very hard to track down. Linux follows an optimistic memory allocation strategy. This means that when malloc() returns non-NULL there is no guarantee that the mem- ory really is available. In case it turns out that the system is out of memory, one or more processes will be killed by the infamous OOM killer. GNU
1993-04-04 MALLOC(3)
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