Question

i have some code:

char * itoa(int a)
{
    char (*t)[16]=(char(*)[16])malloc(1*sizeof(char[16]));
    sprintf(*t,"%d",a);
    return *t;
}
// ...
mvwprintw(my_menu_win,i+1,2,itoa(i));

can i free memory from malloc, without add temporary variables?
e.g:

temp=itoa(i);
mvwprintw(my_menu_win,i+1,2,temp);
free(temp);

Answer 1

In short: No.

To free allocated memory you need a reference to it.

If you could change your conversion API, a possible work around could be to use an externally provided buffer:

char * itoa(char * t, int i)
{
  sprintf(t,"%d",a);
  return t;
}

Call itoa() this way then:

{
  char buffer [16];

  mvwprintw(my_menu_win,i+1,2,itoa(buffer, i));
}

Alternatively (C99 only) one could do the call to itoa() this way:

mvwprintw(my_menu_win,i+1,2,itoa((char[16]){0}, i));

So to clean up this a macro helps:

#define ITOA_0(i) itoa((char[16]){0}, i) /* init array with 0s */
#define ITOA(i) itoa((char[16]){}, i) /* do not init array with 0s -> faster, but none ISO */

...

mvwprintw(my_menu_win,i+1,2,ITOA(i));