Menu
I'm sure it doesn't, but maybe there's black magic in it, so here's my question:
If I have a struct like this:
struct mystr {
char * strp,
unsigned int foo,
};
and I allocate memory for it and want to release it later. Do I have to do
free(mystr_var->strp);
free(mystr_var);
or is the last line enought, does the free() function follow the pointers and free them two?
602
The variable holding the memory address is owned by the process and call to free the memory is actually 'call by value' (the value of the pointer variable is passed), so the free function cannot clear the pointer variable.
free() just declares, to the language implementation or operating system, that the memory is no longer required.
The function free takes a pointer as parameter and deallocates the memory region pointed to by that pointer. The memory region passed to free must be previously allocated with calloc , malloc or realloc . If the pointer is NULL , no action is taken.
Yes, when you use a free(px); call, it frees the memory that was malloc'd earlier and pointed to by px. The pointer itself, however, will continue to exist and will still have the same address. It will not automatically be changed to NULL or anything else.
No, free doesn't follow pointers, you need both lines.
I usually write a function like:
void freemystr(mystr *mystr_var)
{
if (mystr_var)
{
free(mystr_var->strp);
mystr_var->strp = NULL;
free(mystr_var);
}
}
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
