C: malloc error-pointer being freed was not allocated -

i trying return array of string function , free memory used. code below:

int main(int argc, const char * argv[]) {     (int m = 0; m < 10000; m++) {         char **data = datatest();          int = 0;         while(data[i]) {             printf("%p ",data[i]);             free(data[i]);             i++;         }         printf(" address= %p.\n",data);         free(data);     }      return 0; } 

here function:

char **datatest() {     char *row[] = {"this", "is", "a", "data", "string", null};     char **str = row;     char **datareturn = (char **) malloc(sizeof(char *) * 6);      int = 0;     while (*str) {         datareturn[i] = malloc(sizeof(char) * strlen(*str));         strcpy(datareturn[i++], *str);         str++;     }      return datareturn; } 

it runs in beginning, error occurs. below result. address goes wrong somehow , malloc error happens. has met same problem before?

 0x100300030 0x100300040 0x100300050 0x100300060 0x100300070  address= 0x100300000. 0x100300030 0x100300040 0x100300050 0x100300060 0x100300070  address= 0x100300000. 0x100400030 0x100300030 0x100300040 0x100300050 0x100300060  address= 0x100400000. testc(562,0x7fff73e71310) malloc: *** error object 0x3000000000000:      pointer being freed not allocated *** set breakpoint in malloc_error_break debug 0x100300060 0x100300070 0x100300030 0x100300040 0x100300050 0x3000000000000                  program ended exit code: 9 

you need add before return datareturn; in datatest function:

datareturn[i] = null ; 

otherwise while (data[i]) {} continue further wanted.

and instead of:

datareturn[i] = malloc( sizeof(char) * (strlen(*str)) ); 

write:

datareturn[i] = malloc(strlen(*str) + 1); 

in order allocate space terminating zero.

btw sizeof (char) 1.