timer - UNIX 编程。 struct timeval 如何打印它(C 编程)

Question

我正在尝试打印 timeval 类型的值。实际上我可以打印它，但我收到以下警告:

该行有多个标记

• 格式“%ld”需要“long int”类型，但参数 2 的类型为“struct timeval”

程序编译并打印值，但我想知道我是否做错了什么。谢谢。

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

其中的用法是类型

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;

Answer 1

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.

所以你需要这样做

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

获得“格式良好”的时间戳，例如1.000123。