Go中的strconv.Atoi(基本计算器)

问题描述

我试图在Go中制作一个基本的加法计算器(此处是完整的菜鸟)，但是每次输出为0时.

这是代码:

package main

import (
    "fmt"
    "strconv"
    //"flag"
    "bufio"
    "os"
)

func main(){
     reader := bufio.NewReader(os.Stdin)
     fmt.Print("What's the first number you want to add?: ")
     firstnumber, _ := reader.ReadString('\n')
     fmt.Print("What's the second number you want to add?: ")
     secondnumber, _ := reader.ReadString('\n')
     ifirstnumber, _ := strconv.Atoi(firstnumber)
     isecondnumber, _ := strconv.Atoi(secondnumber)
     total := ifirstnumber + isecondnumber
     fmt.Println(total)

}

推荐答案

bufio.Reader.ReadString() 返回直到分隔符为止的数据.因此，您的字符串实际上实际上是"172312 \ n" . strconv.Atoi()不喜欢它并返回0.它实际上返回一个错误，但是您可以通过 _ 忽略它.

bufio.Reader.ReadString() returns data up until and including the separator. So your string actually ends up being "172312\n". strconv.Atoi() doesn't like that and returns 0. It actually returns an error but you're ignoring it with _.

您可以看到此示例:

package main

import (
    "fmt"
    "strconv"
)

func main(){
     ifirstnumber, err := strconv.Atoi("1337\n")
     isecondnumber, _ := strconv.Atoi("1337")
     fmt.Println(err)
     fmt.Println(ifirstnumber, isecondnumber)
}

您可以使用 strings.Trim(number，"\ n")修剪换行符. .

You can trim the newlines with strings.Trim(number, "\n").

这篇关于Go中的strconv.Atoi(基本计算器)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！