c - How are char ** objects saved? -

today, asked me, how char ** objects saved in memory or in binary files. tested following code snippet:

char **array  = (char *)malloc(3 * sizeof(char *));  array[0] = "foo";         // length: 3 array[1] = "long string"; // length: 11 array[2] = "bar";         // length: 3   => full length: 17  int length = 17; 

i write array file:

file *file = fopen(...); fwrite(array, length, 1, file); fclose(file); 

the great thing when read again array file following code, string lengths read without saving more 17 bytes.

file *file   = fopen(...); int   length = 17;  char **array = (char *)malloc(length);  int index        = 0; int parsedlength = 0; while (parsedlength < length) {     char *string       = array[index];     int   stringlength = strlen(string);      printf("%i: \"%s\" (%i)\n", index, string, stringlength);      parsedlength += stringlength;     ++index; } 

i output equals:

0: "foo" (3) 1: "long string" (11) 2: "bar" (3) 

how compiler know, how long each string in array is?

pointers saved file numbers (or better addresses) fact code works illusion created fact saving , loading them inside same program in same run (so since these string literals, address fixed in data segment).

what doing wrong because array's layout in memory following (i'm assuming 4 bytes pointers):

xxxxxxxx yyyyyyyy zzzzzzzz ( = 12 bytes)    ^        ^        ^    |        |       pointer "bar"    |      pointer "long string"  pointer "foo" 

but saving 17 bytes onto file, 12 of memory addresses.

to correctly save strings in file need store whole data contained , length. like:

for (int = 0; < array_length; ++i) {   char *string = array[i];   unsigned int length = strlen(string);   fwrite(&length, sizeof(unsigned int), 1, out);   fwrite(string, sizeof(char), length, out); }  (int = 0; < array_length; ++i {   unsigned int length;   fread(&length, sizeof(unsigned int), 1, in);   array[i] = calloc(sizeof(char), length);   fread(array[i], sizeof(char), length, in); }