The 'main()' question!

I'm intensively trying to learn how to program C++. I am using Visual C++ 2005 Express.

I have had numerous books that have told me to write this:

Code:

---

#include

using namespace std;

main()
{

cout << "Hello World!" << endl;

}

---

Okay, I understand it all, but I don't understand the main() part. Everytime I try to compile it, I get an error. The only way I can get it to compile is if I type int main() or void main(void).

If anyone can explain why this is, it would be more than helpful. Thank-You!

You need to show the return type for main which is int main. Look a Salem's avatar for guidance.

>I have had numerous books that have told me to write this
They were wrong, or too old to be useful. The language standard defines main to return an integer. As such, void main() is illegal and you shouldn't use it. Likewise, the standard has also removed a feature inherited from C called implicit int. That's where if you omit the type in a declaration where implicit int applies, the compiler will assume you meant int. That's why things like this used to work:

Code:

---

main()
{
}

---

However, even when it wasn't illegal, it was a poor practice. The correct definition of main is:

Code:

---

int main()
{
}

---

or:

Code:

---

int main ( int argc, char *argv[] )
{
}

---

You can also use anything equivalent, so explicitly saying void for the argument list is legal as well for main taking no arguments:

Code:

---

int main ( void )
{
}

---

Also note that main is special in how it handles return values. Something like implicit int, if you omit a return statement, 0 for success is returned automagically. Some people prefer to be explicit, which is also fine and the following is equivalent to the above:

Code:

---

int main()
{
  return 0;
}

---

>Everytime I try to compile it, I get an error.
Good, your compiler is correct.

Couldn't be explained any better. :) I feel a bunch of warm fuzzies because I now understand. Thank-You Very Much!

...yet, I am a little bit irritated that my books lied to me. Nonetheless, I am still very giddy.

>...yet, I am a little bit irritated that my books lied to me.
They might not have lied. It could be that you have outdated books. Anything published prior to 2000 is going to be hit or miss for standards compliance.

so what does this mean?

Code:

---

int main ( int argc, char *argv[] )
{
}

---

The first int parameter is the number of arguments from the command line. The second parameter is an array of C style strings containing the arguments. If your program uses command line arguments, then you use that version.

Quote:

---

Originally Posted by lasher

so what does this mean?

Code:

---

int main ( int argc, char *argv[] )
{
}

---

---

For example if you make an application called "app" and you called it with:

app -A -B -C

the argc will become 3 and argv[0] = "-A", argv[1] = "-B", argv[2] = "-C".

Quote:

---

the argc will become 3 and argv[0] = "-A", argv[1] = "-B", argv[2] = "-C".

---

Not exactly
argv[0] = "app"
so argc will be 4 and argv[1] = "-A", argv[2] = "-B", argv[3] = "-C".

Quote:

---

Originally Posted by vart

Not exactly
argv[0] = "app"
so argc will be 4 and argv[1] = "-A", argv[2] = "-B", argv[3] = "-C".

---

Not exactly, again.

The contents of the elements of argv[0] are implementation defined. In practice, in this example, the content will often be as you describe, but could be different. For example, argv[0] will be "app" with a lot of compilers/systems, but it could be "/some/path/app" (where /some/path identifies the directory the app executable is in), or it could simply be an empty string "".