英文:
Optimal way to get time in UTC given today's time as minutes in golang
问题
我有一段代码，需要将以分钟为单位的时间（即小时*60+分钟）转换为time.Time类型。以下是我尝试实现的代码。
func MinutesToTime(minutes int) time.Time {t := time.Now().UTC() // 可以缓存h, m, _ := t.Clock()diff := minutes - (h*60 + m)t = t.Add(time.Duration(diff) * time.Minute)return t}
我的疑问是：
time包是否提供了任何可以帮助我优化这段代码的函数？- 我是否应该直接使用
gettimeofday()系统调用？ - 尝试实现vDSO是否过于复杂？只有amd64架构的vDSO实现可以在Go源代码中找到。
注意：由于分钟是最小的粒度，59秒的误差是可以接受的。
英文:
I have time of the same day in minutes (ie, hours * 60 + minutes) as input ,need to convert it into time.Time here is my attempt to do the same .
Example
Input: 780Output : 2017-01-29 13:00:51.992871217 +0000 UTC
Code
func MinutesToTime(minutes int) time.Time {t := time.Now().UTC() //may be cachedh, m, _ := t.Clock()diff := minutes - (h*60 + m)t = t.Add(time.Duration(diff) * time.Minute)return t}
Doubt
- Is there any functions that
timepackage exposes that might help
me optimize this - Should i be using gettimeofday() system call directly
- Would trying to implement vDSO be an overkill , vDSO implementation for only amd64 is found in go source code
PS
The 59 sec err is acceptable as minute is my lowest granularity
答案1
得分: 1
package mainimport ("fmt""time")func main() {fmt.Print(MinutesToTime(780).UTC())}func MinutesToTime(m int64) time.Time {return time.Unix(time.Now().Unix()/86400*86400+m*60, 0)}
% go run test.go2017-01-29 13:00:00 +0000 UTC%
package mainimport ("fmt""time")func main() {fmt.Print(MinutesToTime(780).UTC())}func MinutesToTime(m int64) time.Time {return time.Unix(time.Now().Unix()/86400*86400+m*60, 0)}
% go run test.go2017-01-29 13:00:00 +0000 UTC%
英文:
package mainimport ("fmt""time")func main() {fmt.Print(MinutesToTime(780).UTC())}func MinutesToTime(m int64) time.Time {return time.Unix(time.Now().Unix()/86400*86400+m*60, 0)}% go run test.go2017-01-29 13:00:00 +0000 UTC%
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