RE: [Dev-C++] (no subject)

[Ioannis V. <no...@ya...>]

argv is an array of C-string pointers which has stored the arguments
given in command line. argv[0] is the name of the program itself. argc
is the number of argv elements and is at least 1 (the name of the
executable). **argv is also *argv[]. The second is the more notationally
correct (because argv is an array of c-strings). So here is some ANSI
C++ 1998 code demonstrating this:
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#include <iostream>
#include <cstdio>
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int main(int argc, char *argv[])
{
     using namespace std;
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     cout<<"The total number of arguments including the filename of the
program,\nargc: "<<argc<<endl;
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     cout<<"\nAnd now the arguments only:\n";
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     for(int i=3D1; i<argc; i++)
        cout<<argv[i]<<" ";
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     cout<<endl;
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     getchar();
}

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How i executed it:
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C:\c>temp 1 test
The total number of arguments including the filename of the program,
argc: 3
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And now the arguments only:
1 test
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Hope you got it.
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Ioannis
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* Ioannis Vranos
* Programming pages: http://www.noicys.f2s.com
<http://www.noicys.f2s.com/>=20
* Alternative URL: http://run.to/noicys
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-----Original Message-----
From: dev...@li...
[mailto:dev...@li...] On Behalf Of
Joh...@ao...
Sent: Friday, November 23, 2001 3:47 AM
To: dev...@li...
Subject: [Dev-C++] (no subject)


#include <iostream.h>
#include <stdio.h>


//QUESTION: Using what feature on the compiler to enter the  'argv'?
void main(int argc, char** argv)
{
     int a =3D 10;
     int b =3D a + argc;
     cout<<" b =3D a + argc: "<<b<<endl;
     int d =3D a + int(argv);//
     cout<<" d =3D a + argv: "<< d<<endl;
    getchar();
}

//I enter the 'argc' through the input of 'Parameter' Button then hit
//the 'Exection' to run the program. But I dont know how/where to enter
the
//'argv' through the command line.....

Very appreciate your help..........=20