文件是否需要在C / C ++中使用open（）打开扩展名？(Does a file need an extension to be opened with open() in C/C++?)

 我正在编写一个使用Xcode for school的程序，需要我们使用open（）系统调用。 我做 
 int input_file_desc = open(input_path, O_RDONLY); 
 printf("input file desc %d:\n", input_file_desc); 
 
 它提出-1。 文件的路径是〜/ data_to_read。 我在xcode中设置了命令行参数。 input_path是我从命令行获得的const char * 。 出于某种原因，如果我将文件名和命令行参数更改为〜/ data_to_read.txt，它可以正常工作。 如果需要更多信息，请与我们联系。 谢谢。 
 
 编辑：我只用.txt尝试它，看看是否是问题，但我仍然不知道为什么它首先需要扩展。 你可以拥有没有扩展名的文件吗？ 在这种情况下它应该仍然有效，只要文件路径和参数都没有扩展名，对吧？ 

I'm writing a program using Xcode for school that requires we use the open() system call. I do
 int input_file_desc = open(input_path, O_RDONLY);
 printf("input file desc %d:\n", input_file_desc);
 
and it comes up with a -1. The file's path is ~/data_to_read. I set up the command line arguments in xcode. input_path is a const char * that i get from the command line. For some reason it works fine if I change the filename and command line argument to ~/data_to_read.txt. Let me know if more info is needed. thanks.
 
EDIT: I only tried it with .txt to see if that was the problem, but I still don't know why it needs an extension in the first place. You can have files without an extension right? In which case it should still work, as long as neither the file path nor the argument has an extension, right?

原文：https://stackoverflow.com/questions/8960297