c - free causing different results from malloc -

below c program have written print different combination of characters in string.

this not efficient way algorithm creats lot of strings. question not how solve problem more efficiently.

the program works(inefficiently though) , prints different combination of characters of string(correctly). when try free strings getting created run issue. free causing issue @ end of thr recur_printc function (it commented).

#include<stdio.h> #include<stdlib.h> #include<string.h>  #define n 3  void recur_printc(char *, int, char *); int main() {     char str[] = "abc";     char *print_arr = malloc(n * sizeof(char));     //call recur_print     recur_printc(print_arr, 0, str);     free(print_arr);     return 0; }  void recur_printc( char *print_arr, int index, char *remaining) {     int i, j, rem_len, index_4_next;     //base case, last cahracter remaining     if(strlen(remaining) == 1)     {         print_arr[index] = remaining[0];         //print print_arr         for(i=0; i<n; i++)         {             printf("%c",print_arr[i]);         }         printf("\n");         return;     }     //if more 1 character remaining     else     {         rem_len = strlen(remaining);         for(i=0; i<rem_len; i++)         {             //add 1 character print_arr             print_arr[index] = remaining[i];             //now create string remaining characters              char *remaining_for_next = malloc((rem_len-1) * sizeof(char));             index_4_next = 0;             for(j=0; j<rem_len; j++)             {                 if(j != i)                 {                     remaining_for_next[index_4_next] = remaining[j];                     index_4_next++;                 }             }             //call recur_print              recur_printc(print_arr, index+1, remaining_for_next);             //free remainin_for_next             /*------this causing issues----*/             //free(remaining_for_next);             remaining_for_next = null;         }     } } 

when ran program in gdb noticed when i=1 first instance of recur_print, strange thing happens malloc.

when line executed :

char *remaining_for_next = malloc((rem_len-1) * sizeof(char)); 

although rem_len-1 equals 2, malloc allocates 3 bytes, , whole algorithm fails beacuse somewhere in code strlen of string used (which 3 instead of 2). not sure happening.(this not happen when comment out free() line.)

below gdb output :

42              char *remaining_for_next = malloc((rem_len-1) * sizeof(char)); (gdb) print remaining_for_next $3 = 0x0 (gdb) n 43              index_4_next = 0; (gdb) print remaining_for_next $4 = 0x602030 "@ `" (gdb) print rem_len-1 $5 = 2 (gdb) q 

sorry long post. once again, question not how print comibnation in different(and better) way. question why above code fails when try free remaining_for_next string (maybe why malloc getting affected).

every time creating string, not appending null terminator, causes error.

so change this:

for(j=0; j<rem_len; j++) {   if(j != i) {     remaining_for_next[index_4_next] = remaining[j];     index_4_next++;   } } 

to this:

for(j=0; j<rem_len; j++) {   if(j != i) {     remaining_for_next[index_4_next] = remaining[j];     index_4_next++;   } } remaining_for_next[index_4_next] = '\0'; 

output:

gsamaras@gsamaras:~/desktop/px$ gcc -wall main.c gsamaras@gsamaras:~/desktop/px$ ./a.out  abc acb bac bca cab cba 

tip: it's must to null terminate strings, not forget it!

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important edit:

as alk noticed, need change this:

char *remaining_for_next = malloc((rem_len - 1) * sizeof(char)); 

to this:

char *remaining_for_next = malloc((rem_len) * sizeof(char)); 

in order make space null terminator.

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_{nice question, +1.}