如何在没有数组的情况下从文件中读取不同大小行的变量

How to read in variables from different size lines from file without an array

本文关键字：读取 变量 数组 情况下 文件      更新时间：2023-10-16

我刚开始编程，不知道很多，但基本上我必须写这个程序：

   Write a program called WeatherStats which will read a RAWS weather file and will return the following statistics:
   Range of dates in the file
   Average high temperature
   Average low temperature
   Maximum high temperature (include date)
   Minimum low temperature (include date)
   Total rainfall
   How many days it rained

我知道如何读取文件并分配我想要的变量，但行不一样，这就是我目前所拥有的：

    if (weatherStats.is_open())
{
        do
        {
            weatherStats >> month >> day >> precip >> hour1 >> hour2 >> temp1 >> temp2 >> hum1 >> hum2 >> elevation >> precipdur1 >> precipdur2;
            cout << month << " " << day << " " << precip << " " << hour1 << " " << hour2 << " " << temp1 << " " << temp2 << " " << hum1 << " " << hum2 << " " << elevation << " " << precipdur1 << " " << precipdur2;
            cout << endl;
        } while (!weatherStats.eof());
    weatherStats.close();
}
system("PAUSE");

}

我遇到的问题是，不是文件中的每一行都有相同数量的数字，所以当我从文件中将它们读入变量时，我在第一行之后得到了错误的数字。

我一直在为此而挣扎，如果有任何帮助，我将不胜感激。

这是文件：

格式：

1. Month 
2. Day 
3. Precipitation – the daily rain amount specified in hundredths of an inch or     millimeters (integer) 
4. Hour 1 – the hour at which the minimum temperature was recorded (0-2400) 
5. Hour 2 – the hour at which the maximum temperature was recorded (0-2400) 
6. Temperature 1 – minimum temperature in degrees Fahrenheit or Celsius (integer) 
7. Temperature 2 – maximum temperature in degrees Fahrenheit or Celsius (integer) 
8. Humidity 1 – maximum humidity in percent, 0 to 99 (integer) 
9. Humidity 2 – minimum humidity in percent, 0 to 99 (integer) 
10. Elevation - feet or meters above sea level 
11. Precipitation Duration (optional) -  the beginning and ending times (0-2400)

7 1 16 2300 1400 50 80 99 26 4570 1700 2000
7 2 0 500 1600 46 84 99 24 4570
7 3 11 400 1500 50 88 99 24 4570 1700 1800
7 4 0 600 1600 54 85 63 28 4570
7 5 0 500 1600 50 76 86 31 4570
7 6 0 500 1600 44 82 83 23 4570
7 7 0 500 1500 43 83 76 14 4570
7 8 0 500 1800 42 84 67 18 4570
7 9 0 500 1600 43 88 69 12 4570
7 10 0 400 1600 46 87 59 14 4570
7 11 0 600 1600 43 76 29 8 4570
7 12 0 400 1700 36 84 51 10 4570
7 13 0 500 1600 39 87 45 10 4570
7 14 0 600 1700 42 86 53 8 4570
7 15 0 200 1500 51 83 44 18 4570
7 16 10 400 1300 50 81 58 22 4570 1100 1200
7 17 0 600 1700 44 84 82 15 4570
7 18 1 500 1200 46 83 77 19 4570 500 600
7 19 0 500 1700 41 87 76 11 4570

您可以使用stringstream。

  • 首先在所有int变量中存储默认值

  • 然后从文件流中读取一行到字符串变量

  • 然后通过stringstream将该字符串放入所有int变量中。

以下是示例代码：（根据您的需要修改）

#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main (){
  //....
  //....
  if (weatherStats.is_open()){
      string str="";
      do{
          month=0;day=0;precip=0;hour1=0;hour2=0;temp1=0;temp2=0;hum1=0;
          hum2=0;elevation=0;precipdur1=0;precipdur2=0;
          //put default values in all variable in each loop
          //NOTE: as I can see in your input file only last two is missing, SO you can here set default values to last two variable only, not for all the variables
          str=weatherStats.ReadLine();
          //read line in to string
          // then put all those in your all variables
          stringstream(str) >> month >> day >> precip >> hour1 >> hour2 >> temp1 >> temp2 >> hum1 >> hum2 >> elevation >> precipdur1 >> precipdur2;
          cout << month << " " << day << " " << precip << " " << hour1 << " " << hour2 << " " << temp1 << " " << temp2 << " " << hum1 << " " << hum2 << " " << elevation << " " << precipdur1 << " " << precipdur2;
          cout << endl;
      } while (!weatherStats.eof());
      weatherStats.close();
  }
  //.......
}

请查看本页的最后部分-字符串流以获取帮助。

这里，若字符串包含较少的数字，那个么额外的变量将不会改变，所以它们的默认值将保持不变。

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