I am from a C background and passing an array in C style causes an error.
package main
import "fmt"
func f(a *int){
fmt.Println(a[1])
}
func main(){
var a [100]int
a[1]=100
f(a)
}
Error:: cannot use a (type [100]int) as type *int in argument to f
As others have mentioned in comments, you probably want to use a slice rather than an array. Slices are passed by reference already, so no need to specify pointers. The make statement below creates a slice of ints (backed by an array). In the code below, I gave it a length of 2 and a capacity of 100 in order to satisfy your goal of assigning to index 1.
import (
"fmt"
)
func f(a []int) {
fmt.Println(a[1])
}
func main() {
a := make([]int, 2, 100)
a[1] = 100
f(a)
}
a inside f...To answer the original question, you can pass a pointer to the array, to pass the array by reference.
For example, the following function changes the contents of an array of 3 integers.
func change(ptr *[3]int) {
ptr[2] = 20
}
You can call this by passing a pointer to the original array:
change(&arr)
package main
import (
"fmt"
)
func main() {
grades := [3]int {1, 2, 3}
fmt.Printf("\nOriginal array:\n")
for i, v := range grades {
fmt.Printf("arr[%d] = %d\n", i, v) // prints 1 2 3
}
change(&grades)
fmt.Printf("\nModified grades:\n")
for i, v := range grades {
fmt.Printf("arr[%d] = %d\n", i, v) // prints 1 2 20
}
}
func change(ptr*[3]int) {
ptr[2] = 20
}
In your example, f is expecting a pointer to an integer,
func f(a *int) // a points to an integer
and you are passing it an array,
var a [100]int
f(a)
hence it gives you the error cannot use a (type [100]int) as type *int in argument to f.
Although passing a pointer to an array is efficient and allows the called function to change the caller’s variable, arrays are still fixed in size. Hence, slices are recommended, which are passed by reference and are dynamic in size.
you must use a slice instead of an array as argument of function. for example:
func p(a int){
fmt.Println(a)
}
func main() {
var a = make([]int,2,10) //2 is length, 10 is capacity
a[0] = 10
p(a[0])
}
array is value type, slice is reference type. you can see more at https://blog.golang.org/go-slices-usage-and-internals
Wish it help you!
The go arrays have limited functionality. In C/C++ when you pass an array to a function the address of the 0th index will be passed but in case of go the complete array will be copied. If you print the address of the array in main and f you can see they will be different if you pass an array. What you can do, is from your array a you can create a slice using a[:]. When slice is passed the value of slice is copied as slice contains pointer to a any changes you do in f will reflect in the original array a.
package main
import "fmt"
func f(a []int){
fmt.Println(a[1])
}
func main(){
var a [100]int
a[1]=100
//Create a slice and pass it
f(a[:])
}
I do not think any of the asweres above works. I found out that i either want to do a global varialbe of type []int or i have to wrap it in custom type like
type wrapper struct {
data []int
}
func f(w *wrapper) {
w.data[0] += 1
w.data = append(w.data, 3)
}
func main() {
w := wrapper{}
w.data = append(w.data, 3)
fmt.Println(w.data)
f(&w)
fmt.Println(w.data)
}
With passing a pointer to my wrapper i was able to avoid passing the size of the array/slice as well as using a global variable.