Linux 将 timeval 结构化为可打印格式
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Struct timeval to printable format
提问by marxin
Could you please help me how to format a struct timevalinstance to human readable format like "2010-01-01 15:35:10.0001"?
你能帮我如何将struct timeval实例格式化为人类可读的格式，比如“2010-01-01 15:35:10.0001”？
采纳答案by Didier Trosset
Convert the tv_secusing localtime, and strftime, then append tv_usecpart.
转换tv_secusing localtime, and strftime，然后附加tv_usec部分。
回答by Frerich Raabe
回答by unwind
You need to manually append the microseconds part, since it's not in the struct tmthat strftime() deals with. Here's a snippet:
您需要手动附加微秒部分，因为它不在struct tmthat strftime() 处理中。这是一个片段：
struct timeval tv;
time_t nowtime;
struct tm *nowtm;
char tmbuf[64], buf[64];
gettimeofday(&tv, NULL);
nowtime = tv.tv_sec;
nowtm = localtime(&nowtime);
strftime(tmbuf, sizeof tmbuf, "%Y-%m-%d %H:%M:%S", nowtm);
snprintf(buf, sizeof buf, "%s.%06ld", tmbuf, tv.tv_usec);
Note how we use explicit precision of 06to get a zero-filled microseconds field. Since the microseconds go from 0 to 999,999, it must always be padded to 6 digits. We don't want to misrepresent e.g. 57 microseconds as 570,000 (compare "1.57" vs "1.000057").
请注意我们如何使用显式精度 of06来获得零填充的微秒字段。由于微秒从 0 到 999,999，因此必须始终填充为 6 位数字。我们不想将例如 57 微秒误表示为 570,000（比较“1.57”与“1.000057”）。
回答by Luo Hongshuai
ctime((const time_t *) &timeval.ts.tv_sec)
I think you are looking for this code, just for your reference.
我认为您正在寻找此代码，仅供参考。
回答by Joe Hildebrand
Combining previous answers and comments, changing the format to be RFC3339-compliant, and checking all of the error conditions, you get this:
结合以前的答案和评论，将格式更改为符合RFC3339并检查所有错误条件，您会得到：
#include <stdio.h>
#include <sys/time.h>
ssize_t format_timeval(struct timeval *tv, char *buf, size_t sz)
{
ssize_t written = -1;
struct tm *gm = gmtime(&tv->tv_sec);
if (gm)
{
written = (ssize_t)strftime(buf, sz, "%Y-%m-%dT%H:%M:%S", gm);
if ((written > 0) && ((size_t)written < sz))
{
int w = snprintf(buf+written, sz-(size_t)written, ".%06dZ", tv->tv_usec);
written = (w > 0) ? written + w : -1;
}
}
return written;
}
int main() {
struct timeval tv;
char buf[28];
if (gettimeofday(&tv, NULL) != 0) {
perror("gettimeofday");
return 1;
}
if (format_timeval(&tv, buf, sizeof(buf)) > 0) {
printf("%s\n", buf);
// sample output:
// 2015-05-09T04:18:42.514551Z
}
return 0;
}
回答by Behrang
Convert the tv_sec using localtime_sinstead of localtime, because if you are writing a global function it may cause some problems. if your function may work in a multi-threaded solution then please consider using localtime_r
使用localtime_s而不是 localtime转换 tv_sec ，因为如果您正在编写全局函数，它可能会导致一些问题。如果您的函数可以在多线程解决方案中工作，那么请考虑使用localtime_r
回答by a3f
This is what I use:
这是我使用的：
#include <time.h>
#include <string.h>
#ifdef _WIN32
#define WIN32_LEAN_AND_MEAN
#include <windows.h>
#include <winsock2.h>
#define gmtime_r(ptime,ptm) (gmtime_s((ptm),(ptime)), (ptm))
#else
#include <sys/time.h>
#endif
#define ISO8601_LEN (sizeof "1970-01-01T23:59:59.123456Z")
char *timeval_to_str(char iso8601[restrict static ISO8601_LEN], unsigned precision, const struct timeval * restrict tv) {
struct tm tm;
if (!gmtime_r(&tv->tv_sec, &tm))
return memcpy(iso8601, "Error: Year overflow", sizeof "Error: Year overflow");
tm.tm_year %= 10*1000;
char *frac = iso8601 + strftime(iso8601, sizeof "1970-01-01T23:59:59.", "%Y-%m-%dT%H:%M:%SZ", &tm);
if (precision) {
unsigned long usecs = tv->tv_usec;
for (int i = precision; i < 6; i++) usecs /= 10;
char *spaces = frac + sprintf(frac - 1, ".%-*luZ", precision, usecs) - 3;
if (spaces > frac) while (*spaces == ' ') *spaces-- = '0';
}
return iso8601;
}
precisionspecifies the width of the seconds fraction. Code is y10k- and yINT_MAX-proof.
precision指定秒部分的宽度。代码是 y10k- 和 y-proof INT_MAX。
