如何从stdin C ++中正确读取和解析整数字符串 [英] How to properly read and parse a string of integers from stdin C++

问题描述

正如标题所说，我正在阅读stdin中的一串整数。我试图读入的数据在文本文件中以下列形式显示：

Just as the title says, I'm reading in a string of integers from stdin. The data i'm trying to read in appears in the following form in a text file:

3
4
7 8 3
7 9 2
8 9 1
0 1 28
etc...    

前两行始终只是单个数字（没有问题！），以下各行各有3个数字。此文件被重定向为stdin到我的程序（myprogram< textfile）。

The first two lines are always just single numbers (no problems there!) and the following lines each have 3 numbers. This file is redirected as stdin to my program (myprogram < textfile).

我尝试了很多东西，但还是没能成功做到这一点！它看起来很简单，但我一直绊倒在哪里（或如何）我应该转换为整数。这是我最近的尝试：

I've tried so many things but have not been able to successfully to do this! It seems so simple but I keep tripped up on where (or how) i should convert to an integer. Here is my latest attempt:

int main()
{

  string str, str1, str2;
  int numCities;
  int numRoads, x, y, z;
  cin >> numCities >> numRoads;

  cout << numCities << " " << numRoads << endl;
  //getline(cin, str1);


  while( getline(cin, str))
  {
    char *cstr;
    cstr = new char[str.size()+1];
    strcpy(cstr, str.c_str());

    x = atoi(strtok(cstr, " ")); //these will be stored in an array or something
    y = atoi(strtok(NULL, " ")); //but right now i just want to at least properly
    z = atoi(strtok(NULL, " ")); //store the appropriate values in these variables!


  }


 return 0;

}

当我尝试使用atoi时，我发生了段错...

I segfault when i try to use atoi...

提前致谢！

推荐答案

首先，您正在使用 cin 读取您的输入，然后使用 getline 。但是，当您阅读最后一个号码时，会丢失'\ n'。你的第一个 getline（cin，str）正在读取'\ n'，这就是你得到段错误的原因。要检查它，请在代码中添加以下行：

First, you are using cin to read your input, then you use getline. However, when you read the last number, there is a '\n' missing. Your first getline(cin, str) is reading the '\n', this is why you get the segfault. To check it, add the following lines to your code:

// just change these lines: 
getline(cin, str);
cout << str << endl;
while(getline(cin, str))
{
    cout << str << endl;

    // ...

您的输出将是：

// output
3 4

7 8 3
7 9 2
8 9 1
0 1 28

您是否只想得到所有数字as int ？对于此输入，我建议：

Do you only want to get all the numbers as int? Than, for this input, I recommend:

while (cin >> x >> y >> z) 
{
    cout << x << " " << y << " " << z;
}

getline（）解决方案

getline() solution

您仍然可以使用 getline 功能。然后，我建议您更改输入读数，以便在前两行使用 getline 。通过这种方式，您将不会遇到'\ n'的问题。但是，我觉得它太复杂了，我不建议。特别是如果它是一个马拉松问题。

You can still use getline function. Then, I suggest you to change your input reading to use getline for your first two lines too. By doing this way, you will not have problems with '\n'. But, I think it gets too complicated and I don't recommend. Specially if it is a marathon problem.

// ...
getline(cin, str);
numCities = atoi( str.c_str() );

getline(cin, str);
numRoads = atoi( str.c_str() );

cout << numCities << " " << numRoads << endl;

while(getline(cin, str))
  {    
    char *cstr;
    cstr = new char[str.size()+1];
    strcpy(cstr, str.c_str());

    x = atoi(strtok(cstr, " "));
    y = atoi(strtok(NULL, " "));
    z = atoi(strtok(NULL, " "));
  }

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