Return local beginning of day time object

EDIT: This only works for UTC times (it was tested in the playground, so the location-specific test was probably wrong). See PeterSO's answer for issues of this solution in location-specific scenarios.

You can use the Truncate method on the date, with 24 * time.Hour as duration:

http://play.golang.org/p/zJ8s9-6Pck

func main() {
    // Test with a location works fine too
    loc, _ := time.LoadLocation("Europe/Berlin")
    t1, _ := time.ParseInLocation("2006 Jan 02 15:04:05 (MST)", "2012 Dec 07 03:15:30 (CEST)", loc)
    t2, _ := time.Parse("2006 Jan 02 15:04:05", "2012 Dec 07 00:00:00")
    t3, _ := time.Parse("2006 Jan 02 15:04:05", "2012 Dec 07 23:15:30")
    t4, _ := time.Parse("2006 Jan 02 15:04:05", "2012 Dec 07 23:59:59")
    t5, _ := time.Parse("2006 Jan 02 15:04:05", "2012 Dec 08 00:00:01")
    times := []time.Time{t1, t2, t3, t4, t5}

    for _, d := range times {
        fmt.Printf("%s\n", d.Truncate(24*time.Hour))
    }
}

To add some explanation, it works because truncate "rounds down to a multiple of" the specified duration since the zero time, and the zero time is January 1, year 1, 00:00:00. So truncating to the nearest 24-hour boundary always returns a "beginning of day".

Both the title and the text of the question asked for "a local [Chicago] beginning of today time." The Bod function in the question did that correctly. The accepted Truncate function claims to be a better solution, but it returns a different result; it doesn't return a local [Chicago] beginning of today time. For example,

package main

import (
    "fmt"
    "time"
)

func Bod(t time.Time) time.Time {
    year, month, day := t.Date()
    return time.Date(year, month, day, 0, 0, 0, 0, t.Location())
}

func Truncate(t time.Time) time.Time {
    return t.Truncate(24 * time.Hour)
}

func main() {
    chicago, err := time.LoadLocation("America/Chicago")
    if err != nil {
        fmt.Println(err)
        return
    }
    now := time.Now().In(chicago)
    fmt.Println(Bod(now))
    fmt.Println(Truncate(now))
}

Output:

2014-08-11 00:00:00 -0400 EDT
2014-08-11 20:00:00 -0400 EDT

The time.Truncate method truncates UTC time.

The accepted Truncate function also assumes that there are 24 hours in a day. Chicago has 23, 24, or 25 hours in a day.