Answer / hello

int main(int argc, char * argv[])
{
//argc has count of no of arguments on command line.
// argv[] contains array of strings seperated by space.
// so argv[1] will give u, second argument, with first
// argument being ur ./a.out(in unix)/test.exe(in win)
// so this way u can access CL args.

// ur code.
}

Answer / rakesh

int main(int argc,char *argv[])
{
int i;
for(i=0;i<argc;i++)
cout<<argv[i];
}

Answer / pb

argc is nothing but Argument Count
argv is nothing Argument Vector (or) Argument Value which
is array of Strings.

The first argument to the Command line is always the
Program Name.

argc contains no. of Arguments that it passed including the
Program Name.So Leave the 1st Argument,

int i;
for(i=1;i<=argc;i++)
cout<<argv[i];

Answer / padmaraj

By using -->
int main(int argc, char * argv[])

In this statement the variable argc holds the at least one
argument of type int , the name of the file it self first
argument and *argv[] is array of char which takes the name
of file.......
If any correction plz E-mail me ...
Bye

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main() { main(); }

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