Golang的类型系统行为-将int除以隐式float

Question

I was experimenting lately with golang's type system and encountered an interesting (or not) behaviour related to floats:

package main

import (
    "fmt"
)

func main() {
    doesntWork()
    works()
}

func doesntWork() {
    x := 1234
    y := x / 100000.0
    s := fmt.Sprintf("%.8f", y)
    fmt.Printf("%s
", s)
}

func works() {
    x := 1234
    y := float64(x) / 100000.0
    s := fmt.Sprintf("%.8f", y)
    fmt.Printf("%s
", s)
}

Go Playground

I think that in the case presented in the example above, in procedure doesntWork(), golang does not automatically "assume" float32/64 when dividing an int by an implicit float? Is that correct? Could anyone point me towards some docs or specs where I could learn more about how golang behaves in situations like the one above?

Answer 1

Here 10000.0 is not float64/32 it is a numeric constant. Go will assume the type of x in this case for the expression and truncate the constant 100000.0 to integer

func doesntWork() {
    x := 1234
    y := x / 100000.0
    s := fmt.Sprintf("%.8f", y)
    fmt.Printf("%s
", s)
}

If you update it to

func doesntWork() {
    x := 1234
    y := x / float64(100000.0)
    s := fmt.Sprintf("%.8f", y)
    fmt.Printf("%s
", s)
}

it will error out saying

 invalid operation: x / float64(100000) (mismatched types int and float64)

Similarly if you change the constant to 100000.111 where truncating to integer will no longer give the same value it will error out

func doesntWork() {
    x := 1234
    y := x / 100000.1
    s := fmt.Sprintf("%.8f", y)
    fmt.Printf("%s
", s)
}

It will error out saying

constant 100000 truncated to integer