How to access the fields of a timeval structure

Question

I'm trying to print the values in a struct timeval variable as follows:

int main()  
{  

    struct timeval *cur;  
    do_gettimeofday(cur);  
    printf("Here is the time of day: %ld %ld", cur.tv_sec, cur.tv_usec);  

    return 0;  
}  

I keep getting this error:

request for member 'tv_sec' in something not a structure or union.  
request for member 'tv_usec' in something not a structure or union.

How can I fix this?

Answer 1

You need to use the -> operator rather than then . operator when accessing the fields. Like so: cur->tv_sec.

Also you need to have the timeval structure allocated. At the moment you are passing a random pointer to the function gettimeofday().

struct timeval cur;
gettimeofday(&cur);
printf("%ld.%ld", cur.tv_sec, cur.tv_nsec);

Answer 2

Because cur is a pointer. Use

struct timeval cur;
do_gettimeofday(&cur);

In Linux, do_gettimeofday() requires that the user pre-allocate the space. Do NOT just pass a pointer that is not pointing to anything! You could use malloc(), but your best bet is just to pass the address of something on the stack.

Answer 3

The variable cur is a pointer of type timeval. You need to have a timeval variable and pass it's address to the function. Something like:

struct timeval cur;
do_gettimeofday(&cur);

You also need

#include<linux/time.h>

which has the definition of the struct timeval and declaration of the function do_gettimeofday.

Alternatively you can use the gettimeofday function from sys/time.h.

Working link

Answer 4

Ok so I tried changing the dot operator to the this operator (->).
I made the change to the declaring the variable to struct timeval cur, not a pointer; and
changed the subsequent call to do_gettimeofday(&cur). Now I have the issue of:
error: storage size of 'cur' isn't known.

Thoughts?

Answer 5

Ok, so I'm running out of linux kernel version 2.6.32, in which I don't believe there is a sys/time.h header, the linux/time.h header contains the definition of the struct timeval
http://lxr.linux.no/#linux+v2.6.32/include/linux/time.h (line 20)

My current version of this main looks like this:

#include <time.h>    
#include stdio.h    
#include unistd.h    

int main()    
{    

struct timeval cur;    
do_gettimeofday(&cur);    
printf("Here's the time of day: %ld.%ld", cur.tv_sec, cur.tv_usec);    

return 0;    
}    

This brings me back to the error of:
storage size of 'cur' is not known

Maybe this will clear something up?