JuniorTheoryCommon

What does the program print?

package main

import "fmt"

type Celsius float64

func (c Celsius) String() string { return fmt.Sprint(float64(c)) + "°C" }

func main() {
    t := Celsius(21)
    fmt.Println(t, float64(t))
}

21°C 21

Expecting 21 21: Celsius has a String() string method, so it satisfies fmt.Stringer, and fmt.Println prints t by calling that method.
Expecting 21°C twice: float64(t) is a plain float64, which has no String method, so fmt prints the bare number 21.

Why does String convert c to float64 before passing it to fmt.Sprint?
Would fmt.Println still call String if the method had the receiver (c *Celsius)?

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Back to questions

JuniorTheoryCommon

What does the program print?

package main

import "fmt"

type Celsius float64

func (c Celsius) String() string { return fmt.Sprint(float64(c)) + "°C" }

func main() {
    t := Celsius(21)
    fmt.Println(t, float64(t))
}

21°C 21

Expecting 21 21: Celsius has a String() string method, so it satisfies fmt.Stringer, and fmt.Println prints t by calling that method.
Expecting 21°C twice: float64(t) is a plain float64, which has no String method, so fmt prints the bare number 21.

Why does String convert c to float64 before passing it to fmt.Sprint?
Would fmt.Println still call String if the method had the receiver (c *Celsius)?

Back to questions Open topic in roadmap →Answer it yourself — Quiz