我正在编写一个套接字程序来维护两个输入套接字的FIFO队列。在决定要服务的队列时,程序从每个队列中提取最新的时间戳。
我需要一种可靠的方法来比较两个timeval
结构。我尝试使用timercmp()
,但我的gcc版本不支持它,文档说明该函数不符合POSIX。
我该怎么办?
答案 0 :(得分:12)
timercmp()
只是libc中的一个宏(sys / time.h):
# define timercmp(a, b, CMP) \
(((a)->tv_sec == (b)->tv_sec) ? \
((a)->tv_usec CMP (b)->tv_usec) : \
((a)->tv_sec CMP (b)->tv_sec))
如果您需要timersub()
:
# define timersub(a, b, result) \
do { \
(result)->tv_sec = (a)->tv_sec - (b)->tv_sec; \
(result)->tv_usec = (a)->tv_usec - (b)->tv_usec; \
if ((result)->tv_usec < 0) { \
--(result)->tv_sec; \
(result)->tv_usec += 1000000; \
} \
} while (0)
答案 1 :(得分:6)
谷歌搜索timeval
给this first result。从该页面:
通常需要减去struct timeval或struct timespec类型的两个值。这是最好的方法。它甚至可以在某些特殊的操作系统上运行,其中tv_sec成员具有无符号类型。
/* Subtract the `struct timeval' values X and Y,
storing the result in RESULT.
Return 1 if the difference is negative, otherwise 0. */
int
timeval_subtract (result, x, y)
struct timeval *result, *x, *y;
{
/* Perform the carry for the later subtraction by updating y. */
if (x->tv_usec < y->tv_usec) {
int nsec = (y->tv_usec - x->tv_usec) / 1000000 + 1;
y->tv_usec -= 1000000 * nsec;
y->tv_sec += nsec;
}
if (x->tv_usec - y->tv_usec > 1000000) {
int nsec = (x->tv_usec - y->tv_usec) / 1000000;
y->tv_usec += 1000000 * nsec;
y->tv_sec -= nsec;
}
/* Compute the time remaining to wait.
tv_usec is certainly positive. */
result->tv_sec = x->tv_sec - y->tv_sec;
result->tv_usec = x->tv_usec - y->tv_usec;
/* Return 1 if result is negative. */
return x->tv_sec < y->tv_sec;
}
答案 2 :(得分:0)
这略有不同,但我认为清楚地说明了所涉及的逻辑。 我正在使用C语言编写一些MSP430代码,并且时间戳结构与timeval非常相似,但是使用nsecs而不是usecs。
此代码保持一切正面,因此无符号整数可以正常工作,并避免溢出(我认为)。它也不会修改传入的时间戳/时间,当然除了结果。
typedef struct timestamp {
int32_t secs;
int32_t nsecs;
} timestamp_t;
int timestamp_sub(timestamp_t * x, timestamp_t * y, timestamp_t * result){
// returns 1 if difference is negative, 0 otherwise
// result is the absolute value of the difference between x and y
negative = 0;
if( x->secs > y->secs ){
if( x->nsecs > y->nsecs ){
result->secs = x->secs - y->secs;
result->nsecs = x->nsecs - y->nsecs;
}else{
result->secs = x->secs - y->secs - 1;
result->nsecs = (1000*1000*1000) - y->nsecs + x->nsecs;
}
}else{
if( x->secs == y->secs ){
result->secs = 0;
if( x->nsecs > y->nsecs ){
result->nsecs = x->nsecs - y->nsecs;
}else{
negative = 1;
result->nsecs = y->nsecs - x->nsecs;
}
}else{
negative = 1;
if( x->nsecs > y->nsecs ){
result->secs = y->secs - x->secs - 1;
result->nsecs = (1000*1000*1000) - x->nsecs + y->nsecs;
}else{
result->secs = y->secs - x->secs;
result->nsecs = y->nsecs - x->nsecs;
}
}
}
return negative;
}
答案 3 :(得分:0)
为了查看时间,我只是简单地说明了这一点。它以字符串形式返回timeval,您可以将其打印或发送到文本文件:
char *tv2str(struct timeval *intv) {
static char ans[200];
snprintf(ans,200,"%u.%u",(unsigned int)intv->tv_sec, \
(unsigned int) intv->tv_usec);
return ans;
}
使用方式:
printf("nowtv: %s\n",tv2str(&nowtv));
nowtv:1568407554.646623
Timercmp()似乎工作不正常,所以我想要一种通过实际查看一些值来检查它的方法。