c - Can I free only a part of a string? -

i filling string of characters , double size time time.

when finish, free unused memory.

void fun (char **str, size_t *len) {   size_t lsi; //last_significant_index   //filling str , reallocating time time.   //*len storing total size of allocated memory @ point    // idea #1   free((*str)[lsi + 1]);     // idea #2   for(size_t = lsi + 1; < *len; i++) {     free(&(*str)[i]);   } } 

none of these ideas seem work however

is possible it? if so, how?

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details:

i using function reallocate strings:

static void increase_list_size(char **list, size_t *list_len) {    size_t new_list_size = (*list_len + 1) * 2; // not allocating list @ declaration, *list_len equals 0.    char *new_list = malloc(sizeof(char) * new_list_size);    (size_t = 0; < *list_len; i++)    {        new_list[i] = (*list)[i];    }    if (list != null) // don't want free empty list (it wasn't allocated @ declaration!    {        free(*list);    }    (*list) = new_list;    *list_len = new_list_size; } 

as can see, allocating two-times more memory every time - that's why wanted free unused memory @ end.

i thought there kind of tricky way it, since felt can use free() free whole memory block.

no, can free() pointers have been returned malloc().

you want use realloc() change allocated memory size smaller (as larger) size. contents of array preserved.

example:

#include <stdlib.h> int main() {     char *str = malloc(100);     ...     str = realloc(str, 50);     ...     free(str); } 

remember check return value of realloc() (as 1 of malloc()) ensure (re)allocation has not failed.