hello

i have a function, that gets a few variables. one of the given variables is a pointer to a string.

prototype:

int add_user(u_elem *list, int conn_num, char *secret)

after executing tthe following line of code:

new=(u_elem *)calloc(sizeof(u_elem),1);

the value in the "secret" variable is lost. it is not just garbage, it is empty. how can I take care that the variable is not going to be cleared?

thanks

raven

IS that second line in the function? IN any case, I think you have the params of calloc backwards. And is new an array?

new is not an array, new is a structure (u_elem).

it looks like this:

typedef struct base
{
char *secret;
int conn_num;
struct base *next;
}u_elem;

I also tryed to allocate memory with malloc instead of calloc, ant then it cut off the contents of the variable secret from the fourth caracter off.

and no, it is not exactly the second line, it is the first line after the variable declarations...

it looks like this:

int i;
u_elem *new;
new=(u_elem *)calloc(sizeof(u_elem),1); <--- this line is now with malloc

thanks

raven

what makes you think that secret should have a value? did you put a value in there to begin with? calloc & malloc init the values to 0 of all memory allocated. so unless you strcpy a value into secret, you'll be missing a value.

AFAIK calloc just returns aligned allocated memory blocks. since you have types of different sizes, you should use malloc

secret is a char*. Before using that I think you should allocate memory for the string that you are going to put there. Else it will have only 4 bytes which the memory allocated for char*.
--Sarin

well, secret has a value, i explicitly give one when calling the function. secret should be set. definitely.

and: ther is memory being allocated for secret. at least in the function where secret is first set. afterwards, I only set pointers. so it should work.

anyway, thanks for the help, I solved one problem, now I am having another. so, see you later in another thread :-)

raven