How do you pass a slice of *interface{} as arguments?

Question

I want to use Scan() in package sql, but the number of columns, and hence the number of arguments, will change at runtime. This is the signature of Scan():

func (rs *Rows) Scan(dest ...interface{}) error

According to the documentation, *interface{} is one of the types accepted by Scan(). So I want to create a slice of []*interface{} and that expand as arguments.

This is what I thought would work:

func query(database *sql.DB) { rows, _ := database.Query("select * from testTable") for rows.Next() { data := make([]*interface{}, 2) err := rows.Scan(data...) // Compilation error fmt.Printf("%v%v\n", *data[0], *data[1]) if err != nil { fmt.Println(err.Error()) } } }

Compilation fails with cannot use data (type []*interface {}) as type []interface {} in argument to rows.Scan. I thought that data... would expand to &data[0], &data[1], but apparently not. I don't understand the error message. *interface{} is compatible with interface{}, so why can't I expand the slice of pointers to interface types?

This works:

func query(database *sql.DB) { rows, _ := database.Query("select * from testTable") for rows.Next() { data := make([]*interface{}, 2) err := rows.Scan(&data[0], &data[1]) // Only changed this line fmt.Printf("%v%v\n", *data[0], *data[1]) // Outputs "[48][116 101 120 116]" if err != nil { fmt.Println(err.Error()) } } }

I can't use this however, because the number of columns is unknown at compile time. How can I write this code so that I can pass a variable number of *interface{} to rows.Scan()?

Why do you want []*interface{} instead of []interface{}? — Jonathan Hall
– Jonathan Hall, Commented Dec 7, 2018 at 8:12
I just thought that []*interface{} gets me closest to the answer. I will edit my question to add this detail. — MakotoE
– MakotoE, Commented Dec 7, 2018 at 16:43

Jonathan Hall · Accepted Answer · 2018-12-07 08:00:38Z

First, you must not use []*interface{} slice of pointers to interface rather than []interface{} where the interfaces are pointers. []*interface{} is different from []interface{}. Just create a slice of interfaces where each element is a pointer to a concrete type.

Here is a snippet how you would do this.

var x int var s string data := []interface{}{&x, &s} rows.Scan(data...)

Note on the use of the ... spread operator.

Here are some related questions that will explain a bit more:

golang: slice of struct != slice of interface it implements?

Cannot convert []string to []interface {}

calvinsomething · Accepted Answer · 2022-09-27 16:02:25Z

If you really want to pass a []*interface{} (perhaps you don't know the concrete types of the output) you must first wrap each *interface{} in a interface{}:

values := make([]interface{}, columnsCount) for i := range values { values[i] = new(interface{}) }

Individual values passed into a ...interface{} parameter are automatically wrapped in a interface{}, but just like []int... won't satisfy ...interface{}, neither will []*interface{}....

This works great if you don't know anything about the query. Thanks!

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