如何使用命令行参数打开 C++ 中的文本文件，而不使用文件扩展名？

Question

I'm writing a C++ program where I pass the name of a text file as a command-line argument and then manipulate that text file. But I'm having trouble referencing the text file. The thing is, I want to be able to reference the file passed in just by its name without the extension. For example I want to be able to reference the code like this:

./ProgramName exampleTextFileName

and not like this:

./ProgramName exampleTextFileName.txt

I'm able to access the file by simply opening the file name stored in argv[1] and using.txt on the command line. But how would I do this without having to pass in.txt at the end? I tried doing this by taking argv[1] and manually adding quotation marks and.txt, but I get an error when I try to open the file using the appended name. I assume the variable type for file names is not actually a string? How would I do this correctly?

Here is the code I was trying to use:

int main(int argc, char *argv[]) {
string line;
string fileName;
fileName = argv[1];
fileName = "\"" + fileName + ".txt\"";
cout << fileName;
ifstream myfile (fileName);
if (myfile.is_open()) {
    while ( getline (myfile,line) ) {
        cout << line << '\n';
    }
myfile.close();
}
else cout << "Unable to open file"; 


return 0;}

Answer 1

You don't need double quotes for fileName , this should be enough:

...
string fileName = argv[1];
fileName += ".txt";
...

I'd would also add a check on argc to make sure an argument was actually given.