What's the order of this Go print statements?

Question

I'm taking 'A Tour of GO' tutorial about Golang and this code:

package main

import (
    "fmt"
    "math"
)

func pow(x, n, lim float64) float64 {
    if v := math.Pow(x, n); v < lim {
        return v
    } else {
        fmt.Printf("%g >= %g\n", v, lim)
    }
    // can't use v here, though
    return lim
}

func main() {
    fmt.Println(
        pow(3, 2, 10),
        pow(3, 3, 20),
    )
}

would print "27 >= 20 9 20".

I'm rather confused why it's not "9 27 >= 20 20"

Shouldn't the first call to pow(3,2,10) return 9, print it, then call pow(3,3,20) and print the rest?

score 7 · Accepted Answer · answered Sep 16 '15 at 21:31

This is actually somewhat subtle and had me confused for a second. The "secret" is that a function must evaluate all its arguments before it's called. So it calls the function twice to get 9 and 20, but one of those evaluations happens to call Println.

It's fairly straightforward why the language evaluates its arguments before calling a function (that sort of partial application is tricky when side effects are involved, and is mostly reserved for functional languages), but hiding functions with side effects such as printing inside a function evaluation should probably be discouraged just for the purposes of clarity.

The code is perhaps more straightforwardly:

func main() {
    arg1,arg2 := pow(3,2,10),pow(3,3,20)
    fmt.Println(arg1, arg2)
}

score 4 · Answer 2 · answered Sep 16 '15 at 21:46

fmt.Println() needs all its parameters before it gets invoked.

So, you have 9, but while you're getting the other value (20) the console prints something else.

Is just like this:

package main

import "fmt"

func main() {

    fmt.Println(9, f())
}
func f() int {
    fmt.Println("This gets printed before")
    return 20
}

Output:

This gets printed before
9 20

What's the order of this Go print statements?

2 Answers2