c - permutations for pass generation -

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from original code generates possible combinations of 3 numbers:

void gen(char *word, char *chars, int ug, int length) {     size_t = 0;     int u = ug;     while(i < strlen(chars)){         word[u] = chars[i++];          if(u < (length-1)) {             gen(word, chars, ++u, length);             u--;         } else {             printf("%s\n", word);         }     } }  int main(int argc, char* argv[]) {      char *chars = "0123456789";     /* 3 char long */     int = 3;      int length = 3;     /* allocate memory output */     char *word = calloc(length, 0);      gen(word, chars, 0, 3);     return 0; }

but cause need function work differently, modified this:

char *genpass(char* pass,int len, int crt, size_t i) {     char *chars = "0123456789";     pass[crt] = chars[i++];     return pass; }   int main(int argc, char* argv[]) {     char *pass = calloc(10, 0);     int crt;//current permutation     int len = 3;     size_t = 0;     (crt=0;crt<10;crt++)     {         pass = genpass(pass,len,crt,i);         printf("pass: %s\n", pass);         i++;         //some other code work pass     }     return 0; }

but returns:

pass: 0 pass: 01 pass: 012 pass: 0123 pass: 01234 pass: 012345 pass: 0123456 pass: 01234567 pass: 012345678 pass: 0123456789

what did mess up? how can make generate correctly first 10 permutations of 3 length numbers?

now function genpass() not recursive!

if want generate 10 permutations, in code when printf , increment counter each time printf permutation, , when counter 10 break while


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