Unpack slices on assignment?

As Sergio Tulentsev mentioned, general packing/unpacking as is done in Python is not supported. I think the way to go there is to define your own small ad-hoc function using multiple return values:

func splitLink(s, sep string) (string, string) {
    x := strings.Split(s, sep)
    return x[0], x[1]
}

And you can then write:

name, link := splitLink("foo\thttps://bar", "\t")

But this will obviously work only when at least two substrings are being split, and silently ignore if more than two were. If this is something you use a lot, it might make your code more readable though.

--EDIT--

Another way to unpack an array is via variadic pointer arguments:

func unpack(s []string, vars... *string) {
    for i, str := range s {
        *vars[i] = str
    }
}

Which let you write:

var name, link string
unpack(strings.Split("foo\thttps://bar", "\t"), &name, &link)
bookmarks[name] = link

This will work for any array size, but it is arguably less readable, and you have to declare your variables explicitly.

If your function is meant to split a string only by the first occurrence of the separator, you can always make your own function:

package main

import (
    "fmt"
    "strings"
)

func Split(s, sep string) (string, string) {
    // Empty string should just return empty
    if len(s) == 0 {
        return s, s
    }   

    slice := strings.SplitN(s, sep, 2)

    // Incase no separator was present
    if len(slice) == 1 {
        return slice[0], ""
    }

    return slice[0], slice[1]
}

func main() {
    a, b := Split("foo;bar;foo", ";")
    fmt.Println(a, b)
}

Output:

foo bar;foo

Playground

You could also use anonymous functions:

a, b := func() (string, string) {
    x := strings.Split("foo;bar", ";")
    return x[0], x[1]
}()

Note: don't forget the () on the end of the closing bracket } otherwise you will get the error:

assignment mismatch: 2 variable but 1 values

This is because without the () a function (1 value) is returned not the expected strings (2 values).