C++ 如何根据用户输入打开文件？
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How to open file based on user input?
提问by user1725435
This code keeps giving me errors i dont know what Im doing wrong? How can I request the user to input the filename and then open that file and perform tasks with the data.
这段代码不断给我错误，我不知道我做错了什么？如何请求用户输入文件名，然后打开该文件并使用数据执行任务。
Example:
例子：
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int main ()
{
ifstream inData;
ofstream outData;
string fileName;
cout << "Enter the data file Name : " << endl;//asks user to input filename
cin >> fileName; //inputs user input into fileName
inData.open(fileName); //heres where i try to open the file with the users input?
outData.open(fileName);
cout << fileName;
return 0;
}
The code keeps giving me errors. I have tried to use getline? I want the fileName to be string and not char.
代码不断给我错误。我试过用getline吗？我希望文件名是字符串而不是字符。
回答by 0x499602D2
You're passing std::stringobjects to the parameters, not actual null-terminated char pointers. To do that, use c_str:
您将std::string对象传递给参数，而不是实际的以空字符结尾的字符指针。为此，请使用c_str：
inData.open(fileName.c_str());
outData.open(fileName.c_str());
