calling golang println with variable arguments

Question

I have the following code to print n lines for debug purposes. output() prints the address of args instead of the parameters. How to fix it ?

var outputMax = 10
var outputCnt = 0

func output(args ...interface{}) {
    outputCnt++
    if(outputCnt < outputMax) { println(args) }
}

func main() {
    for i := 0; i < 5; i++ {
        output("Value of i is now:", i)
    }
}

score 4 · Accepted Answer · answered Dec 03 '13 at 14:07

The usual way to call a varargs function would be like so:

func output(args ...interface{}) {
    println(args...)
}

However, this will give you a invalid use of ... with builtin println compilation error. If you switch to fmt.Println() instead, it should work.

calling golang println with variable arguments

1 Answers1