Computer Science Canada Command Line Arguments |
| Author: | LearningC++ [ Sun Nov 29, 2009 10:21 pm ] |
| Post subject: | Command Line Arguments |
I'm using a tutorial and I got to command line arguments. I don't get some of the code: #include <fstream> #include <iostream> using namespace std; int main ( int argc, char *argv[] ) { if ( argc != 2 ) // argc should be 2 for correct execution cout<<"usage: "<< argv[0] <<" <filename>\n"; else { ifstream the_file ( argv[1] ); if ( !the_file.is_open() ) cout<<"Could not open file\n"; else { char x; while ( the_file.get ( x ) ) cout<< x; } } } Why does argc have to be 2 for it to be correct? What is this program trying to do? |
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| Author: | Tony [ Sun Nov 29, 2009 10:32 pm ] |
| Post subject: | RE:Command Line Arguments |
The correct usage of this program requires the argument counter to be 2: program name + filename. Try running the program to see what it does (or read the source code). |
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| Author: | LearningC++ [ Sun Nov 29, 2009 10:45 pm ] |
| Post subject: | Re: Command Line Arguments |
Running it displays the location of the program. |
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| Author: | Tony [ Sun Nov 29, 2009 10:58 pm ] |
| Post subject: | RE:Command Line Arguments |
as part of the "usage: " message? |
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| Author: | LearningC++ [ Sun Nov 29, 2009 11:25 pm ] |
| Post subject: | RE:Command Line Arguments |
Ya argv[0] is just the filename I guess. |
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| Author: | DtY [ Mon Nov 30, 2009 6:32 pm ] |
| Post subject: | Re: RE:Command Line Arguments |
LearningC++ @ Sun Nov 29, 2009 11:25 pm wrote: Ya argv[0] is just the filename I guess. argv[0] is what ever you used to start the program, so if you rub ./program it will show different output than if you ran /full/path/to/program |
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