4 ms·

It is perfectly understandable from my perspective, but the caveat is that I'm used to Go's syntax and idioms and they are quite different from the usual. It h

• zvrba 15y ago
  Can you explain the " ; true" part?
  • chrisjsmith 15y ago
    Straight from the tutorial: "Since the switch value is just true, we could leave it off—as is also the situation in a for statement, a missing value means true. In fact, such a switch is a form of if-else chain. While we're here, it should be mentioned that in switch statements each case has an implicit break." The basic outcome is that: 1. The assignment to er, nr is an initialization statement for the switch. 2. The true (default value if not specified) is used to configure the switch as an if-else chain which is required as the assignment above makes the purpose of the switch ambiguous (is the result of the assignment configuring the switch - how do you do that as multiple values are returned?). You could rewrite it: nr, er := f.Read(buf[:]); switch true { ... } or even: nr, er := f.Read(buf[:]); switch { ... } But the switch initializer scopes it to the switch block cleanly.
    • zvrba 15y ago
      "switch true {...}" or "switch {...}"? really? So how would switch 3.141592654 { ... } affect the case-statements inside?
      • chrisjsmith 15y ago
        No idea as I wouldn't do it :) (sorry - cop out that)
      • supersillyus 15y ago
        When you say "switch <value>", it matches the "case" statements based on if the value after "case" is equal to <value>. So, if you say const P = 3.141 switch 3.141 { case P: fmt.Println("This prints") } When no value is specified after "switch", the value of "true" is implied, which is why you can do: switch { case a && b: .. case something(): }