C ++可以在Code :: Blocks中以完整路径打开文件，但不能以相对路径作为命令行参数

Question

I am working on a C++ homework project in Code::Blocks where the file name to open is specified in the first command line argument. The file I am trying to open is titled InputFile1.txt. When I run the code below:

#include <iostream>
#include <cstdlib>
#include <fstream>
#include <unistd.h>
using std::cout;
using std::ifstream;
using std::ofstream;
using std::cerr;

int main(int argc, char *argv[])
{
     ifstream inputFile;

     puts(getcwd(0,1234));

     inputFile.open(argv[1]);
     if (inputFile.is_open())
     {
         cerr << "Couldn't open file " << argv[1];
         return(EXIT_FAILURE);
     }

     cout << "\nThe file to search is: " << argv[1];

     return(EXIT_SUCCESS);
}

The error message "Couldn't open file InputFile1.txt" is displayed. However when I type in the full path to InputFile1.txt the file successfully opens as indicated by the output of the message on the bottom. The following troubleshooting steps have been taken:

1. Successfully opened the file with a C program running on the same IDE.
2. Verified the file name was correctly spelt and was the first command line argument; also opened the file with gedit.
3. Verified the file path matched the current working directory output by the puts(getcwd(0,1234) statement.
4. Verified the Code::Blocks project execution working directory (Projects > Properties > Build targets - Execution working dir:) matched the output of step 3, and that InputFile1.txt was in the working directory.
5. Changed the working directory to a directory one level up and copied InputFile1.txt to the directory. Once again, success with the full path but not as argv[1].

My analysis:

• I don't think it is file permissions or else it wouldn't open in gedit or with the full path.
• I don't think it is the Execution working directory or it wouldn't have opened for the C program run inside of Code::Blocks.
• It is something to do with my C++ code or Code::Blocks configuration.

Any insights/help would be greatly appreciated.

Answer 1

It turns out that the code listed above was working correctly. The if statement in the code below:

 inputFile.open(argv[1]);
 if (inputFile.is_open())
 {
     cerr << "Couldn't open file " << argv[1];
     return(EXIT_FAILURE);
 }

was evaluating the result of the expression inputFile.is_open() , which returned true when the file was successfully opened. To check if the file had not been successfully opened, the expression !inputFile.is_open() should have been used. Thanks to my prof for helping me resolve this issue - the error was all my own.