标签：call ror OLE this The 很多 any converter rest
原题目：
Your friend is traveling abroad to the United States so he wrote a program to convert fahrenheit to celsius. Unfortunately his code has some bugs.
Find the errors in the code to get the celsius converter working properly.
To convert fahrenheit to celsius:
celsius = (fahrenheit - 32) * (5/9)Remember that typically temperatures in the current weather conditions are given in whole numbers. It is possible for temperature sensors to report temperatures with a higher accuracy such as to the nearest tenth. Instrument error though makes this sort of accuracy unreliable for many types of temperature measuring sensors.
大致意思是，需要将温度做个转换，把华氏的温度转成摄氏的温度，题目给出了一个公式
celsius = (fahrenheit - 32) * (5/9)
输入一个华氏温度，通过函数计算判断是否高于冰点（0℃）,题目给出一个写好的代码，需要将代码优化后才能提交。
给的代码的bug有很多，运行的时候各种坑，感觉是一个新手菜鸟会犯的很多错。（提交了答案后找不回来了，不列举）
按照题目意图写了如下代码，但是在提交的是，有部分被否定了。
被否代码：
1 def weather_info (temp): 2 c =(temp-32)*5/9 3 if (c > 0): 4 return (str(c) + " is above freezing temperature") 5 else: 6 return (str(c) + " is freezing temperature")
#‘252.77777777777777 is above freezing temperature‘ should equal ‘252.7777777777778 is above freezing temperature‘
查看了其他的解决方法后，发现自己在做计算的时候没有将c =(temp-32)*5/9 的数字类型转换程float ,导致精度损失，被否。
修改了代码后提交成功了。
成功提交代码
1 def weather_info (temp): 2 c =(temp-32.0)*(5.0/9.0) 3 if (c > 0): 4 return (str(c) + " is above freezing temperature") 5 else: 6 return (str(c) + " is freezing temperature")
-------我是分割线
题目给的代码找回来了，标红的都是有问题的代码。
1 def weather_info (temp): 2 c : convert(temp) 3 if (c > 0): 4 return (c + " is freezing temperature") 5 else: 6 return (c + " is above freezing temperature") 7 8 def convertToCelsius (temperature): 9 var celsius = (tempertur) - 32 + (5/9)
codewar刷题--8kyu--Grasshopper - Debug
标签：call ror OLE this The 很多 any converter rest
原文地址：https://www.cnblogs.com/zivaro/p/13334665.html
