UNIX Programming. struct timeval how to print it (C-programming)

I am trying to print a value of type timeval. Actually I am able to print it, but I get the following warning:

Multiple markers at this line

format ‘%ld’ expects type ‘long int’, but argument 2 has type ‘struct timeval’

The program compiles and it prints the values, but I would like to know if I am doing something wrong. Thanks.

    printf("%ld.%6ld\n",usage.ru_stime);
    printf("%ld.%6ld\n",usage.ru_utime);

where usage is of type

typedef struct{
    struct timeval ru_utime; /* user time used */
    struct timeval ru_stime; /* system time used */
    long   ru_maxrss;        /* maximum resident set size */
    long   ru_ixrss;         /* integral shared memory size */
    long   ru_idrss;         /* integral unshared data size */
    long   ru_isrss;         /* integral unshared stack size */
    long   ru_minflt;        /* page reclaims */
    long   ru_majflt;        /* page faults */
    long   ru_nswap;         /* swaps */
    long   ru_inblock;       /* block input operations */
    long   ru_oublock;       /* block output operations */
    long   ru_msgsnd;        /* messages sent */
    long   ru_msgrcv;        /* messages received */
    long   ru_nsignals;      /* signals received */
    long   ru_nvcsw;         /* voluntary context switches */
    long   ru_nivcsw;        /* involuntary context switches */
}rusage;

struct rusage usage;

Solution

In the GNU C Library, struct timeval:

is declared in sys/time.h and has the following members:
long int tv_sec
This represents the number of whole seconds of elapsed time.
long int tv_usec
This is the rest of the elapsed time (a fraction of a second), represented as the number of microseconds. It is always less than one million.

So you will need to do

printf("%ld.%06ld\n", usage.ru_stime.tv_sec, usage.ru_stime.tv_usec);

to get a "nicely formatted" timestamp like 1.000123.