Please paste text as text, not as an image. Thanks.

This is expected behavior. In floating point arithmetic, dividing a finite value by zero gives you an infinity, not an error. See "Floating-point operators" at https://golang.org/ref/spec#Arithmetic_operators.

package main

import (
	"testing"
)
func TestConstInt(t *testing.T){
	const x int =1
	t.Log(x/0)
}
func TestVarInt(t *testing.T){
	var x int =2
	t.Log(x/0)
}
func TestVarFloat64(t *testing.T){
	var x float64 =2.222
	t.Log(x/0)
}
func TestConstFloat64(t *testing.T){
	const x float64 =2.222
	t.Log(x/0)
}
func TestConstComplex(t *testing.T){
	const x complex128=complex(1,2)
	t.Log(x/0)
}
func TestVarComplex(t *testing.T){
	var x complex128=complex(1,2)
	t.Log(x/0)
}

This is the test code I wrote. Why use const to modify float64 and complex will cause errors.
@ianlancetaylor

In Go, constant arithmetic does not use the same rules as variable arithmetic. In particular, constant floating point values have no infinity, or NaN, or negative zero. See https://golang.org/ref/spec#Constant_expressions .

In general for questions about the language it's better to use a forum. You will get better and faster answers. See https://golang.org/wiki/Questions.