Should true or false terminate callback iteration? - go

In some languages it's necessary or cleaner to do iteration by providing a callback function that receives items and returns a boolean that indicates whether to continue or stop the iteration.
Which is the preferred value to indicate desire to stop/continue? Why? What precedents exist?
Example in Go:
func IntSliceEach(sl []int, cb func(i int) (more bool)) (all bool) {
for _, i := range sl {
if !cb(i) {
return false
}
}
return true
}

Which is the preferred value to indicate desire to stop/continue?
true for continue
Why?
Example 1:
func example(i interface{}) {
if w, ok := i.(io.Writer); ok {
// do something with your writer, ok indicates that you can continue
}
}
Example 2:
var sum int = 0
it := NewIntStatefulIterator(int_data)
for it.Next() {
sum += it.Value()
}
In both cases true (ok) indicates that you should continue. So I assume that it would be way to go in your example.

Foreword: The following answer applies to a callback function which decides based on the current item(s) whether the loop should terminate early - this is what you asked.
This is not to be confused with a function that progresses and reports if there are more elements to process, where a true return value is generally accepted to signal that there are more elements (for which a good example is Scanner.Scan()), and whose typical use is:
scanner := bufio.NewScanner(input)
for scanner.Scan() {
// Process current item (line):
line := scanner.Text()
fmt.Println(line) // Do something with line
}
Sticking to bool return type
Usually returning true to indicate termination results in code that is easier to read. This is due to the nature of for: if you do nothing, for continues, so you have to explicitly break if you want to terminate early, so having a clean termination condition is more common.
But it's a matter of taste. You may go whichever you like, but what's important is to name your callback function in a meaningful way that will clearly state what its return value means, and thus looking at the code (the condition in which it is used) will be easily understandable.
For example the following names are good and the return value is unambiguous:
// A return value of true means to terminate
func isLast(item Type) bool
func terminateAfter(item Type) bool
func abort(item Type) bool
// A return value of true means to continue (not to terminate)
func keepGoing(item Type) bool
func carryOn(item Type) bool
func processMore(item Type) bool
Using these results in easily understandable code:
for i, v := range vals {
doSomeWork()
if terminateAfter(v) {
break // or return
}
}
for i, v := range vals {
doSomeWork()
if !keepGoing(v) {
break // or return
}
}
// Or an alternative to the last one (subjective which is easier to read):
for i, v := range vals {
doSomeWork()
if keepGoing(v) {
continue
}
break
}
As negative examples, the following callback function names are bad as you can't guess what their return value mean:
// Bad: you can't tell what return value of true means just by its name:
func test(item Type) bool
func check(item Type) bool
Having error return type
It's also common for the callback to not just test but also do some work with the passed item. In these cases it is meaningful to return an error instead of a bool. Doing so, obviously the nil return value indicates success (and to continue), and a non-nil value indicates error and that processing should stop.
func process(item Type) error
for i, v := range vals {
if err := process(v); err != nil {
// Handle error and terminate
break
}
}
Having enum-like return value
Also if multiple return values have meaning, you may choose to define constants for return values, which you can name meaningfully.
type Action int
const (
ActionContinue Action = iota
ActionTerminate
ActionSkip
)
func actionToTake(item Type) Action
for i, v := range vals {
switch actionToTake(v) {
case ActionSkip:
continue
case ActionTerminate:
return
}
doSomeWork()
}

Related

Return default value for generic type

How do you return nil for a generic type T?
func (list *mylist[T]) pop() T {
if list.first != nil {
data := list.first.data
list.first = list.first.next
return data
}
return nil
}
func (list *mylist[T]) getfirst() T {
if list.first != nil {
return list.first.data
}
return nil
}
I get the following compilation error:
cannot use nil as T value in return statement
You can't return nil for any type. If int is used as the type argument for T for example, returning nil makes no sense. nil is also not a valid value for structs.
What you may do–and what makes sense–is return the zero value for the type argument used for T. For example the zero value is nil for pointers, slices, it's the empty string for string and 0 for integer and floating point numbers.
How to return the zero value? Simply declare a variable of type T, and return it:
func getZero[T any]() T {
var result T
return result
}
Testing it:
i := getZero[int]()
fmt.Printf("%T %v\n", i, i)
s := getZero[string]()
fmt.Printf("%T %q\n", s, s)
p := getZero[image.Point]()
fmt.Printf("%T %v\n", p, p)
f := getZero[*float64]()
fmt.Printf("%T %v\n", f, f)
Which outputs (try it on the Go Playground):
int 0
string ""
image.Point (0,0)
*float64 <nil>
The *new(T) idiom
This has been suggested as the preferred option in golang-nuts. It is probably less readable but easier to find and replace if/when some zero-value builtin gets added to the language.
It also allows one-line assignments.
The new built-in allocates storage for a variable of any type and returns a pointer to it, so dereferencing *new(T) effectively yields the zero value for T. You can use a type parameter as the argument:
func Zero[T any]() T {
return *new(T)
}
In case T is comparable, this comes in handy to check if some variable is a zero value:
func IsZero[T comparable](v T) bool {
return v == *new(T)
}
var of type T
Straightforward and easier to read, though it always requires one line more:
func Zero[T any]() T {
var zero T
return zero
}
Named return types
If you don't want to explicitly declare a variable you can use named returns. Not everyone is fond of this syntax, though this might come in handy when your function body is more complex than this contrived example, or if you need to manipulate the value in a defer statement:
func Zero[T any]() (ret T) {
return
}
func main() {
fmt.Println(Zero[int]()) // 0
fmt.Println(Zero[map[string]int]()) // map[]
fmt.Println(Zero[chan chan uint64]()) // <nil>
}
It's not a chance that the syntax for named returns closely resembles that of var declarations.
Using your example:
func (list *mylist[T]) pop() (data T) {
if list.first != nil {
data = list.first.data
list.first = list.first.next
}
return
}
Return nil for non-nillable types
If you actually want to do this, as stated in your question, you can return *T explicitly.
This can be done when the type param T is constrained to something that excludes pointer types. In that case, you can declare the return type as *T and now you can return nil, which is the zero value of pointer types.
// constraint includes only non-pointer types
func getNilFor[T constraints.Integer]() *T {
return nil
}
func main() {
fmt.Println(reflect.TypeOf(getNilFor[int]())) // *int
fmt.Println(reflect.TypeOf(getNilFor[uint64]())) // *uint64
}
Let me state this again: this works best when T is NOT constrained to anything that admits pointer types, otherwise what you get is a pointer-to-pointer type:
// pay attention to this
func zero[T any]() *T {
return nil
}
func main() {
fmt.Println(reflect.TypeOf(zero[int]())) // *int, good
fmt.Println(reflect.TypeOf(zero[*int]())) // **int, maybe not what you want...
}
You can init a empty variable.
if l == 0 {
var empty T
return empty, errors.New("empty Stack")
}

How to combine two return params into one value

What I have many times in different versions in my code:
func f() (bool, bool) {
value, successFulOperation := someStuff()
return value, successFulOperation
}
// somewhere else
value, successfulOperation := f()
if value && successfulOperation {
// do stuff
}
// do stuff should only be executed if value is true AND the operation that retrieved value succeeded without an error. In other words: I don't care about value or successfulOperation. I only care about value && successfulOperation.
A solution I want to avoid (seems verbose):
value, successfulOperation := f()
actualValue := value && successfulOperation
if actualValue {...}
Above code is really simplified. In reality the if conditions will be nested and more complicated.
What I want:
A wrapper for f that combines both values into one. How do I do that? The solution should work for any function taking any parameters and returning two bools.
The following does not work:
type doubleBoolFunc func(...interface{}) (bool, bool)
func and(fn doubleBoolFunc, params ...interface{}) bool {
b1, b2 := fn(params...)
return b1 && b2
}
actualValue := and(f())
You can't write a wrapper function to turn the two bools into one until generics are in the language in go 1.181. Or at least you can, using reflect, but it's a mess.
But you can write this with go 1.17:
func both(x, y bool) bool {
return x && y
}
func f() (bool, bool) {
return true, false
}
func main() {
r := both(f())
fmt.Println(r)
}
But more practical (in my opinion) is to eschew this complication, and use a 1-line if. Not everything needs to be a function or abstracted away:
if a, b := f(); a && b {
...
}
[1] Even when generics are introduced in go 1.18, I don't think there'll be a way to specific a generic type that represents a function with arbitrary arguments that returns two bools.
func and(a, b bool) bool {
return a && b
}
then if f return 2 bool
value := and(f())
will work
To fix the code at the end of your question, don't invoke the function f(), simply reference the function name f and then list its args:
// actualValue := and(f(args...))
actualValue := and(f, args...)
https://go.dev/play/p/KAT2L58ZQy3

What is the difference between named return value and normal return value?

My question is about the different named return value vs normal return value.
my code
package main
import "fmt"
func main() {
f := fmt.Println
f(a())
f(b())
}
func a() int {
i := 0
defer func() {
i += 1
fmt.Println("a defer : ", i)
}()
return i
}
func b() (i int) {
i = 0
defer func() {
i += 1
fmt.Println("b defer : ", i)
}()
return i
}
the result:
the a function return 0
the b function reutrn 1
Why?
The named return value also allocates a variable for the scope of your function.
func a() int: While you already return the value of i = 0, but since no named values was defined the static value got returned. So even though you're increasing i in the deferred function it doesn't affect the returned value.
func b() (i int): The variable i is allocated (and already initialized to 0). Even though the deferred function runs after the i = 0 was returned the scope is still accessible and thus still can be changed.
Another point of view: you can still change named return values in deferred functions, but cannot change regular return values.
This especially holds true in the following example:
func c() (i int) {
defer func() {
i = 1
fmt.Println("c defer : ", i)
}()
return 0
}
defer runs a function after the return statement but before the function is auctually returned, thus enabling modify returned results. However, only named return results can be accessed normally, i.e. by the variable name.
The return statement, when not naked (another thing about named return, but irrelevant here), the expression got evaluated. And if the return is named, the named variable is assigned with the evaluated value.
In your code, in func a() int the return is typed but not named. So when return i is execuated, it sets the return value, a variable not available to the code, as the value of i. You can consider it as RETVAL := i. And later, your deferred function modified i but the return value (RETVAL) remains unchanged.
But in func b() (i int), i is a named return. Thus, when return i execuate, it literally translate to i = i. And later, your deffered function modified i, a return value, so the returned value change.
More on return: https://golang.org/ref/spec#Return_statements
With the named return value you directly modify what gets returned, with the "normal" return value you just modify local variable in the scope of your function, which never gets returned.
More on this:
Deferred function can access named return values but it has no return value itself - so this is actually the only way to modify main function results from there. Very useful thing.
Imagine you want to fix code which panics - you want it to return the error instead. You can solve it by using recover in a deferred function and then assigning recovered error to named return value.
Example, somewhat abstract but hopefully useful:
func noMorePanics() (err error) {
defer func() {
if r := recover(); r != nil {
err = r.(error)
}
}()
potentiallyPanickingFunction()
}

golang comparing two structs that are implemented differently

I am new to go language, under learning. I have years OOP experience in C++. There is a stacker interface that is written in go and two implementations of it, one is slice base stack and another one is linkedlist base.
I find it is hard to compare two different structures and tell if they are containing the same data or not. The simple example code list below (notice a lot of functions/implementations are not listed because they are not relative with this question). The key function is stackEquals, I have tried different ways to approach it but they failed. Please see the comments in the code.
package main
import (
"fmt"
"errors"
)
// The interface is fixed, cannot be modified
type Stacker interface {
isEmpty() bool
size() int
push(x int)
peek() (int, error)
pop() (int, error)
copy() Stacker
}
type StackSlice struct {
slice []int
}
type StackLinked struct {
next *StackLinked
value int
// possible with other variables that is not relative
}
// There are interface function/method implementations did not paste
func (s StackSlice) String() string {
// return all the value inside the stack as string
// like [5 4]
}
func (s StackLinked) String() string {
// return all the value inside the stack as string
// like [5 4]]
}
// Pre-condition:
// none
// Post-condition:
// returns true if s and t have the same elements in the same order;
// both s and t have the same value after calling stackEquals as before
// Annoying constraint:
// Use only Stackers in the body of this functions: don't use arrays,
// slices, or any container other than a Stacker.
func stackEquals(s, t Stacker) bool {
// This implementation below always return false unless they are the same thing
return s == t
// I tried return s.String() == t.String() but gave an error said interface doesn't have String() method.
}
How can I compare two stacks that implemented in different way and tell if they are the same (same means same values in same order) in the stack.
If you need to compare two interfaces, you can only use the methods in that interface, so in this case, String does not exist in the interface (even though both of your implementations have it, the interface itself does not).
A possible implementation would be:
func stackEquals(s, t Stacker) bool {
// if they are the same object, return true
if s == t {
return true
}
// if they have different sizes or the next element is not the same,
// then they are different
if s.size() != t.size() || s.peek() != t.peek() {
return false
}
// they could be the same, so let's copy them so that we don't mess up
// the originals
ss = s.copy()
tt = t.copy()
// iterate through the values and check if each one is
// the same. If not, return false
for ; i, err := ss.pop(); err == nil {
if j, err := tt.pop(); err != nil || i != j {
return false
}
}
return true
}
This assumes that the only error pop would get is when there are no more values, otherwise you will need to do some better error checking and use isEmpty.

how to make function support one or two return value

in go tutorial following code is often seen:
a := foo()
b, c := foo()
or actually what I see is:
m["Answer"] = 48
a := m["Answer"]
v, ok := m["Answer"]
how many foo() is defined?
Is it two, one with one return type, another with two return type?
Or just one foo() with two return type defined, and somehow magically when only need one return value (a := foo()), another return value is omitted?
I tried
package main
func main() {
a := foo()
a = 1
}
func foo() (x, y int) {
x = 1
y = 2
return
}
func foo() (y int) {
y = 2
return
}
But I got error message foo redeclared in this block
While some built in operations support both single and multiple return value modes (like reading from a map, type assertions, or using the range keyword in loops), this feature is not available to user defined functions.
If you want two versions of a function with different return values, you will need to give them different names.
The Effective Go tutorial has some good information on this.
Basically, a function defines how many values it returns with it's return statement, and it's function signature.
To ignore one or more of the returned values you should use the Blank Identifier, _(Underscore).
For example:
package main
import "fmt"
func singleReturn() string {
return "String returned"
}
func multiReturn() (string, int) {
return "String and integer returned", 1
}
func main() {
s := singleReturn()
fmt.Println(s)
s, i := multiReturn()
fmt.Println(s, i)
}
Playground
The v, ok := m["answer"] example you've given is an example of the "comma, ok" idiom (Also described in the Effective Go link above). The linked documentation uses type assertions as an example of it's use:
To extract the string we know is in the value, we could write:
str := value.(string)
But if it turns out that the value does not contain a string, the program will crash with a run-time error. To guard against that, use the "comma, ok" idiom to test, safely, whether the value is a string:
str, ok := value.(string)
if ok {
fmt.Printf("string value is: %q\n", str)
} else {
fmt.Printf("value is not a string\n")
}
If the type assertion fails, str will still exist and be of type string, but it will have the zero value, an empty string.

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