Passing arguments to function...

i'm using the arguments to main and then passing them to another function to open a file with the argument as the file name...this is the code:

Code:

---

int main(int argc, char* argv[])
{
    char *fname;
    fname = argv[1];
    readFile(fname);
}

---

this is the readFile function:

Code:

---

void readFile(char *file)
{
    ifstream fin (file);

}

---

it doesn't work...what am i doing wrong? i've only given the pertinent code...

Regards,

Farooq

You need an actual ifstream variable. You should also make sure to check that you have the correct number of command line arguments:

Code:

---

#include <fstream>
#include <iostream>

void readFile(char *filename)
{
    std::ifstream input(filename);
    // Do stuff with input file stream
}

int main(int argc,char *argv[])
{
    if( argc != 2 )
    {
        std::cerr << "Error: Incorrect number of command-line arguments." << std::endl;
        return 1;
    }
    else readFile(argv[1]);
    return 0;
}

---

still doesn't work...it hangs on execution...didn't i do pretty much the same thing in my code?

thanks

Farooq

#1) place a cout << fname << endl; after the line fname = argv[1]; to be sure you have valid information in fname.

#2) you need to declare an ifstream object in order to associate with a file using the () operator, say maybe.

ifstream fin(file)

rather than

ifstream(file);

then be sure fin opened file before trying to use it.

ok i'm sorry about that ive changed the original post...that is how i've done it in my original code...ok 2 things...

1) does argv[1] indicate the first or second argument to main...i mean when i run my program (astar) i want to write:

astar inputfile

where it is understood that inputfile is the name of the file...so is that argv[0] or argv[1]...i tried argv[0] and i get a segmentation error...argv[1] doesn't work as it hangs...

2) if i hardcode the filename it always works...e.g. if i write:

Code:

---

ifstream fin ("inputfile.txt");

---

this always works...i just don't want to hardcode the name of the file...

Regards,

Farooq

argv[1] should be inputfile in your example. check it out by running the program with the command line arguments you have listed. comment out the call to readfile() and put a cout << argv[1] << endl; in the program to prove it. Put in a cout << fname << endl; after you assign argv[1] to fname to be sure you have the same information int fname.

You'll need to type in astar inputfile.txt and not astar inputfile when you run the program if you want to mimic that hardcoded ifstream fin("inputfile.txt"); piece of code.

ya it works when i use:

astar inputfile.txt

is there someway where i can just make it: astar inputfile

i really don't know if it will be a text file or what...in fact the assignment question just says it will be run as:

astar inputfile

Regards,

Farooq

you could provide code to append the .txt string on to argv[1].

you could ask user to enter extension type of the file and append it to the end of argv[1].