When will ofstream::open fail?

Question

I am trying out try, catch, throw statements in C++ for file handling, and I have written a dummy code to catch all errors. My question is in order to check if I have got these right, I need an error to occur. Now I can easily check infile.fail() by simply not creating a file of the required name in the directory. But how will I be able to check the same for outfile.fail() (outfile is ofstream where as infile is ifstream). In which case, will the value for outfile.fail() be true?

sample code [from comments on unapersson's answer, simplified to make issue clearer -zack]:

#include <fstream> using std::ofstream;  int main()  {     ofstream outfile;     outfile.open("test.txt");     if (outfile.fail())          // do something......      else          // do something else.....      return 0;  } 

Answer 1

The open(2) man page on Linux has about 30 conditions. Some intresting ones are:

• If the file exists and you don't have permission to write it.
• If the file doesn't exist, and you don't have permission (on the diretory) to create it.
• If you don't have search permission on some parent directory.
• If you pass in a bogus char* for the filename.
• If, while opening a device file, you press CTRL-C.
• If the kernel encountered too many symbolic links while resolving the name.
• If you try to open a directory for writing.
• If the pathname is too long.
• If your process has too many files open already.
• If the system has too many files open already.
• If the pathname refers to a device file, and there is no such device in the system.
• If the kernel has run out of memory.
• If the filesystem is full.
• If a component of the pathname is not a directory.
• If the file is on a read-only filesystem.
• If the file is an executable file which is currently being executed.