Golang passing pointer to sql .Scan() returns pointers

Question

Confused as to why the below is happening. TLDR; I'm passing a pointer and a "&val" to database/sql 's Scan method, and getting different results

rows, err := db.Query(selectRowsQuery)
val = new(int)

for rows.Next() {
  err = rows.Scan(val)
...
fmt.Println(val)      // 0xc0000b6c48

Val's value is a pointer. Why does val stay as a pointer?

but if I do this:

var val int
...
err = rows.Scan(&val)
fmt.Println(val)          // 0

val returns as the value. Aren't "&val" (val is an int value) and "val" (val is a pointer to int) the same thing?

To print the integer instead of the pointer in the first snippet of code, dereference the pointer: fmt.Println(*val) — user5728991
– user5728991, Commented Aug 24, 2020 at 4:50
Please take the Tour of Go for language fundamentals or consult any C reference on what pointers are and how they behave. — Volker
– Volker, Commented Aug 24, 2020 at 5:37

Sudhir Jonathan · Accepted Answer · 2020-08-24 06:28:00Z

2

In the first case, when you do val = new(int) you are getting a pointer, and rows.Scan(val) is putting a value into the location that pointer is pointing to correctly, but you're printing fmt.Println(val): which just means you want Go to print the pointer. Which it did... the string representation of a pointer is the address it's pointing to. What you'd want to do is fmt.Println(*val), which says 'print the value that the pointer is pointing to'.

The second example works correctly because var val int initializes a value, and fmt.Println(val) prints it correctly. err = rows.Scan(&val) required you to send the address of val by calling &val because that's how we tell rows.Scan where to put the data instead of passing it data that it can't do anything with.

answered Aug 24, 2020 at 6:28

Sudhir Jonathan

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1 Comment

webbyweb Over a year ago

Thanks!! Very clear explanation, makes sense now :)

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