为什么从这个简单的Go代码中得到这样的输出？

Question

I am new to Go and very confused with one of the sample codes in the tutorial. Here is the code and its output:

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
", v, lim)
    }
    return lim
}

func main() {
    fmt.Println(pow(3, 3, 10), pow(4, 3, 20))
}

When I run it, the output of this code is:

27 >= 10
64 >= 20
10 20

Which is quite strange to me! I expect this output:

27 >= 10
10
64 >= 20
20

Can someone please help me to understand what is going on with Println in this code?

Answer 1

In Go, as in the vast majority of modern programming languages, the arguments of a function call are evaluated before the function call itself. Consider the following function:

func add(i, j int) int {
  return i + j
}

And let's call it like this:

func a() int {
  return 2
}

func b() int {
  return 3
}

// ... somewhere in main()
add(a(), b())

You would expect that, by the time the code of add runs, the values of i and j are known, right? The call would then return 5 as expected. To enable this, the Go compiler arranges for the arguments of add - namely a() and b() to be run before add itself runs.

---

The same happens in your code sample with Println and pow. The arguments to Println need to be evaluated before its body is run, and evaluating pow causes other things to be printed.