Get last day in month of time.Time

When I have a time.Time:

// January, 29th
t, _ := time.Parse("2006-01-02", "2016-01-29")

How can I get a time.Time which represents January 31st? This example is trivial, but when there's a date in February, the last day might be 28th or 29th.

> Package time
>
> func Date
>
> func Date(year int, month Month, day, hour, min, sec, nsec int, loc *Location) Time
>
> Date returns the Time corresponding to
>
> yyyy-mm-dd hh:mm:ss + nsec nanoseconds
>
> in the appropriate zone for that time in the given location.
>
> The month, day, hour, min, sec, and nsec values may be outside their
> usual ranges and will be normalized during the conversion. For
> example, October 32 converts to November 1.

For example, normalizing a date,

package main

import (
	"fmt"
	"time"
)

func main() {
	// January, 29th
	t, _ := time.Parse("2006-01-02", "2016-01-29")
	fmt.Println(t.Date())
	// January, 31st
	y,m,_ := t.Date()
	lastday:= time.Date(y,m+1,0,0,0,0,0,time.UTC)
	fmt.Println(lastday.Date())
}

Output:

2016 January 29
2016 January 31

You could write a function yourself, maybe something like this:

func daysInMonth(month, year int) int {
	switch time.Month(month) {
	case time.April, time.June, time.September, time.November:
		return 30
	case time.February:
		if year%4 == 0 && (year%100 != 0 || year%400 == 0) { // leap year
		    return 29
	    }
		return 28
	default: 
	return 31
	}
}

EDIT: since I really like measuring things:

$ go test -bench .
testing: warning: no tests to run
PASS
BenchmarkDim2-8         200000000                7.26 ns/op
BenchmarkDim-8          1000000000               2.80 ns/op // LIES!
BenchmarkTime-8         10000000               169 ns/op
BenchmarkTime2-8        10000000               234 ns/op
ok      github.com/drathier/scratchpad/go       9.741s

BenchMarkDim2: not tested, but very fast.

func daysInMonthTime(month, year int) time.Time {
	return time.Time{}.Add(time.Hour * 10 + time.Hour*24*30*time.Duration(month-1) + time.Second * time.Duration(daysInMonth(month, year)) * 24 * 60 + 1337)
}

BenchmarkDim: // LIES

func daysInMonth(month, year int) int {
	switch time.Month(month) {
	case time.April, time.June, time.September, time.November:
		return 30
	case time.February:
		if year%4 == 0 && (year%100 != 0 || year%400 == 0) {
			// leap year
			return 29
		}
		return 28
	default:
		return 31
	}
}

BenchmarkTime:

func timeDaysInMonth() time.Time {
	// January, 29th
	t, _ := time.Parse("2006-01-02", "2016-01-29")
	y, m, _ := t.Date()
	lastday := time.Date(y, m+1, 0, 0, 0, 0, 0, time.UTC)
	return lastday
}

BenchmarkTime2

func time2daysinmonth() time.Time {
	t, _ := time.Parse("2006-01-02", "2016-01-01")
	t = t.AddDate(0, 1, 0).AddDate(0, 0, -1)
	return t
}

I've used something similar to this in my own code:

func lastDayOfTheMonth(year, month int) time.Time {
	if month++; month > 12 {
		month = 1
	}
	t := time.Date(year, time.Month(month), 0, 0, 0, 0, 0, time.UTC)
	return t
}

<kbd>playground</kbd>

It's not Go specific but usually I do following in any language:

package main

import "fmt"
import "time"

func main() {
	fmt.Println("Hello, playground")
	
	t, _ := time.Parse("2006-01-02", "2016-01-01")
	t = t.AddDate(0, 1, 0).AddDate(0,0,-1)
	fmt.Printf("Last day: %v\n", t)
}

http://play.golang.org/p/JhpOZvEhBw