使用Golang中的类型断言检测值超出范围的错误

Question

Given the following code:

iv, err := strconv.ParseInt("18446744073709551448", 10, 64)
fmt.Println(iv)
fmt.Printf("%#v
", err)
fmt.Printf("%v
", err)

//Output:
9223372036854775807
&strconv.NumError{Func:"ParseInt", Num:"18446744073709551448", Err:(*errors.errorString)(0x1040a040)}
strconv.ParseInt: parsing "18446744073709551448": value out of range

How can I detect that the function failed due to being out of range of an int64? The strconv.ParseInt function returns an error type, but in this case it is actually a strconv.NumError type as indicated by %#v. The Error Handling and Go article mentions you can use type assertion to check for specific types of errors, but it doesn't give any examples. What expression should I use to complete this code:

if expression {
    uv, err := strconv.ParseUint("18446744073709551448", 10, 64)
}

Answer 1

We have,

Package strconv

var ErrRange = errors.New("value out of range")

ErrRange indicates that a value is out of range for the target type.

type NumError struct {
        Func string // the failing function (ParseBool, ParseInt, ParseUint, ParseFloat)
        Num  string // the input
        Err  error  // the reason the conversion failed (ErrRange, ErrSyntax)
}

A NumError records a failed conversion.

func (e *NumError) Error() string

For example,

package main

import (
    "fmt"
    "strconv"
)

func main() {
    iv, err := strconv.ParseInt("18446744073709551448", 10, 64)
    if err != nil {
        if numError, ok := err.(*strconv.NumError); ok {
            if numError.Err == strconv.ErrRange {
                fmt.Println("Detected", numError.Num, "as a", strconv.ErrRange)
                return
            }
        }
        fmt.Println(err)
        return
    }
    fmt.Println(iv)
}

Output:

Detected 18446744073709551448 as a value out of range