Dynamically initialize array size in go

Question

I try to write a small application in go that takes 'x' numbers of integers from standard input, calculates the mean and gives it back. I have only gotten so far:

func main() {
var elems, mean int
sum := 0

fmt.Print("Number of elements? ")

fmt.Scan(&elems)

var array = new([elems]int)

for i := 0; i < elems; i++ {
    fmt.Printf("%d . Number? ", i+1)
    fmt.Scan(&array[i])
    sum += array[i];
}............

When trying to compile this I get the following error message:

invalid array bound elems

What is wrong here?

i saw this somewhere already :) let me guess hackerrank.com/challenges/simple-array-sum/problem? — bin2hex
– bin2hex, Commented Feb 24, 2018 at 20:09

John Smith · Accepted Answer · 2022-04-14 19:14:51Z

167

You should use a slice instead of an array:

//var array = new([elems]int) - no, arrays are not dynamic
var slice = make([]int,elems) // or slice := make([]int, elems)

See "go slices usage and internals". Also you may want to consider using range for your loop:

// for i := 0; i < elems; i++ { - correct but less idiomatic
for i, v := range slice {

edited Apr 14, 2022 at 19:14

John Smith

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answered Dec 16, 2011 at 20:27

Paolo Falabella

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7 Comments

shutefan Over a year ago

Thanks for the link! Exactly what I sought :)

user811773 Over a year ago

@Paolo: You would need to write "for i,_ := range array {}". In my opinion, using "for i:=0; i<elems; i++ {}" seems more appropriate here. In either case, there is no need to use an array in this program.

Paolo Falabella Over a year ago

@Atom: I thought he could use v instead of array[i] in the loop. But I agree that my search for idiomatic code is sometimes exaggerated (I'm one of those people who, when abroad, look for typical restaurants and get typically ripped off)

Ivan Chau Over a year ago

Should be slice := make([]int, elems) to avoid confusion.

alramdein Over a year ago

if it's "dynamic", why we need to set the length of the slice?

|

user811773 · Accepted Answer · 2011-12-17 10:00:47Z

In my opinion, this results from confusion over the usage of the new and make functions. This is a known issue/feature in the Go language, as evidenced by several discussions about new vs make at golang-nuts.

The difference between new and make may become clearer by letting Go print out the type of the value created by new and make:

package main

import "fmt"

func main() {
    fmt.Printf("%T  %v\n", new([10]int), new([10]int))
    fmt.Printf("%T  %v\n", make([]int, 10), make([]int, 10))
}

The output:

*[10]int  &[0 0 0 0 0 0 0 0 0 0]
[]int  [0 0 0 0 0 0 0 0 0 0]

As can be seen from the type, to access an array element of new([10]int) we would first need to dereference the pointer.

Both new and make require a Go type as their 1st argument. However, the expression [elems]int is not a Go type (unless elems is a Go constant, which isn't the case here).

For further reference, see http://golang.org/doc/go_spec.html#Allocation and http://golang.org/doc/go_spec.html#The_zero_value.

To get a better understanding of whether the result of new is usable, it may be helpful to lookup whether len and cap work with zero (nil) values: http://golang.org/doc/go_spec.html#Length_and_capacity

Codefor · Accepted Answer · 2013-12-03 12:13:45Z

4

See The Go Programming Language Specification

http://golang.org/ref/spec#Array_types

http://golang.org/ref/spec#Constants

It says:"The length is part of the array's type; it must evaluate to a non- negative constant representable by a value of type int. "

Constants by no means vary.

answered Dec 3, 2013 at 12:13

Codefor

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