Question

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

此行有多个标记

format'％ld'需要类型'long int'，但参数2的类型为'struct timeval'

程序编译并打印值，但我想知道我做错了什么。感谢。

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

其中使用的类型为

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：

在sys / time.h中声明并且具有以下成员：
long int tv_sec
这表示经过时间的整秒数。
long int tv_usec
这是剩余的经过时间（几分之一秒），表示为微秒数。它总是不到一百万。

所以你需要做

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

获取“格式正确”的时间戳，例如1.000123。

Answer 2

由于struct timeval将被声明为：

struct timeval {
    time_t      tv_sec;
    suseconds_t tv_usec;
}

你需要了解基础领域：

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

Answer 3

是的，timeval定义如下

struct timeval { 
    time_t      tv_sec; 
    suseconds_t tv_usec; 
}

使用

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

肯定会有所帮助。

Answer 4

是的，

int main( void )
{
    clock_t start, stop;
    long int x;
    double duration;
    static struct timeval prev;
    struct timeval now;

    start = clock();  // get number of ticks before loop

    for( x = 0; x < 1000000000; x++ );
    // sleep(100);

    stop = clock();  // get number of ticks after loop

    // calculate time taken for loop
    duration = ( double ) ( stop - start ) / CLOCKS_PER_SEC;

    printf( "\nThe number of seconds for loop to run was %.2lf\n", duration );

    gettimeofday(&now, NULL);
    prev.tv_sec = duration;
    if (prev.tv_sec)
    {
        int diff = (now.tv_sec-prev.tv_sec)*1000+(now.tv_usec-prev.tv_usec)/1000;
        printf("DIFF %d\n",diff);
    }

    return 0;

}

Answer 5

我只是根据上面的信息组成了这个方便的小功能。独立的，只需要时间。h。无论您想知道标准输出流中的时间，都可以使用所需标签来调用它。

void timestamp(char *lbl) { // just outputs time and label
  struct timeval tval;
  int rslt;
  rslt = gettimeofday(&tval,NULL);
  if (rslt) printf("gettimeofday error\n");
  printf("%s timestamp: %ld.%06ld\n", lbl, tval.tv_sec, tval.tv_usec);
}

典型输出如下： dpyfunc得到了ftqmut时间戳：1537394319.501560

而且，您可以在#ifdef周围加上对它的调用，以通过注释掉#define一次打开和关闭所有调用。这几乎像概要分析一样有用，但是您可以快速将其禁用以用于生产/发布代码。喜欢：

#define TIMEDUMPS

#ifdef TIMEDUMPS
timestamp("function 1 start");
#endif

#ifdef TIMEDUMPS
timestamp("function 2 start");
#endif

注释掉#define TIMEDUMPS，然后将其全部关闭。不管有多少源代码文件。

Answer 6

.tv_sec可以是-ve，并且发生这种情况时，.tv_usec有偏差（强制范围为[0..1000000]），因此：

#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>

static void frac(intmax_t usec)
{
    int precision = 6;
    while (usec % 10 == 0 && precision > 1) {
        precision--;
        usec = usec / 10;
    }
    printf(".%0*jd", precision, usec);
}

static void print_timeval(struct timeval tv)
{
    struct timeval d;
    /*
     * When .tv_sec is -ve, .tv_usec is biased (it's forced to the
     * range 0..1000000 and then .tv_sec gets adjusted).  Rather
     * than deal with that convert -ve values to +ve.
     */
    if (tv.tv_sec < 0) {
        printf("-");
        struct timeval zero = {0,};
        timersub(&zero, &tv, &d);
    } else {
        d = tv;
    }
    printf("%jd", (intmax_t)d.tv_sec);
    if (d.tv_usec > 0) {
        frac(d.tv_usec);
    }
}

int main()
{
    for (intmax_t i = 1000000; i > 0; i = i / 10) {
        for (intmax_t j = 1000000; j > 0; j = j / 10) {
            struct timeval a = { .tv_sec = i / 1000000, .tv_usec = i % 1000000, };
            struct timeval b = { .tv_sec = j / 1000000, .tv_usec = j % 1000000, };
            struct timeval d;
            timersub(&a, &b, &d);
            printf("%7jd us - %7jd us = %7jd us | %2jd.%06jd - %2jd.%06jd = %2jd.%06jd | ",
                   i, j, i - j,
                   (intmax_t)a.tv_sec, (intmax_t)a.tv_usec,
                   (intmax_t)b.tv_sec, (intmax_t)b.tv_usec,
                   (intmax_t)d.tv_sec, (intmax_t)d.tv_usec);
            print_timeval(d);
            printf("\n");
        }
    }
    return 0;
}

UNIX编程。 struct timeval如何打印它（C编程）

6 个答案: