free() function without malloc or calloc

问题

quick question

Can you use the free() function without having to prior call a malloc ??

ei.

void someFunc( void )
{
   char str[6] = {"Hello"};

   //some processing here ....

   free(str);
}

I get no compiling errors but Does this work or is it correct at all ?

Thank you,

回答1:

This is not at all correct:

You cannot free a static array such as char str[6].
free() should only be called on memory you allocated (or on NULL).

回答2:

When you call malloc() or any other allocation function, memory will be allocated on the heap. This is the only memory that can be freed. When you declare a static string, as you've done in your example, the string is allocated at compile time in another memory segment. The same goes for the str pointer itself which is allocated on the stack, and thus cannot be freed either.

回答3:

Using free on a non-malloc'd variable will result in a Segfault generally. Example:

#include <stdlib.h>

int main()
{
  char str[6] = {"Hello"};
  free(str);
}

$ gcc test.c -o test

$ ./test

Segmentation fault

回答4:

free() uses data prepended to the allocated block to manage the heap. If the memory pointed to was not allocated by a heap allocation function such as malloc() or calloc(), then the data preceeding the block will be meaningless as heap management data.

Some libraries will detect invalid heap data and yieled a runtime error, otherwise the behaviour is undefined. Often the consequences of such an error will remain unnoticed until you later attempt to allocate further memory. This can make debugging such errors very difficult.

You would not get a compiler error because it is not a syntactic error and is not detectable at compile time. The compiler has no knowledge of the semantics of library functions. All it knows is that malloc() returns a void* and that free() accepts a void*; there is no way of knowing at compile time whether the pointer refers to a dynamically allocated block because the memory is by definition allocated at runtime. Also a pointer may be modified at runtime to point to any memory type, or may be aliased - copied to another pointer and then free'd through the second pointer. You expect a lot of the compiler if you expect an error message; however some static analysis tools may be able to warn if such an error may occur, and dynamic analysis tools such as valgrind may detect the error when and if it actually occurs during testing.

回答5:

No

The free(3) function takes a void * parameter, so you can pass it any sort of pointer without a compile-time error. But bad things will happen if the pointer wasn't originally returned by malloc(3) and never previously given back to free(3).

来源：https://stackoverflow.com/questions/4110767/free-function-without-malloc-or-calloc

标签

malloc

calloc