Free () without malloc or calloc

Question

quick question

Can you use the free () function without calling malloc first ??

e.

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

   //some processing here ....

   free(str);
}

I cannot compile errors, but does this work or is everything correct?

Thank,

Answer 1

This is not entirely correct:

• You cannot free a static array, for example char str[6].
• free () should only be called in allocated memory (or in NULL).

Answer 2

malloc() , . , . , , . str, stack , , .

Answer 3

non-malloc'd Segfault. :

#include <stdlib.h>

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

$gcc test.c -o test

$./test

Answer 4

free() , . , malloc() calloc(), , , .

, undefined. , . .

, . . , , , malloc() void * free() void *; , , . , , - , . , ; , , , valgrind, , .

Answer 5

No

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The function free(3)takes a parameter void *, so you can pass it any pointer without a compile-time error. But bad things will happen if the pointer was not originally returned malloc(3)and never previously returned to free(3).