Meaning of int main( int _argc, char *_argv[] )

Question

3.00/5 (1 vote)

See more:

What does this mean in plain English?

C++

int main( int _argc, char *_argv[] )

I have a small if further down that depends on it and i can't figure it out....

C++

if ((argc>=4) && (strcmp(argv[3], "m")==0))
			more=1;
		//**********************************
		result = atoi(argv[2]);
		packetCount=0;

Posted 20-Aug-11 19:53pm

Member 7766180

Updated 8-Mar-22 19:58pm

Add a Solution

3 solutions

Solution 2

Main is a function that returns int, it takes 2 parameters - the first tells it how many arguments were passed on the command line, the second is an array of pointers to each of the arguments.

I.e argc(ount) and argv(alues)

Posted 20-Aug-11 20:21pm

enhzflep

Comments

Member 7766180 22-Aug-11 10:01am

Thank you...I understand that, but what is the point of this?
if (argc<=2)
{
printf("\nUseage...");
printf("\ndood [IP-address] [packet-count] (ml) (o)");
printf("\n--> eyeball ");
return EXIT_SUCCESS;
}

Solution 3

thanks ur answers helped me alot

Posted 21-Aug-11 19:27pm

arishmass

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Sergey Alexandrovich Kryukov · Accepted Answer · 2011-08-20T20:05:00

Solution 1

In plain English, _argc means number of arguments in the command line used at the call of the application, _argv is the array of string arguments in the command line.

The code sample is a bit weird as only arguments #2 and #3 are used; and #0 and #1 ignored. Maybe this is an incomplete sample.

This is how Microsoft explains command line: http://msdn.microsoft.com/en-us/library/17w5ykft.aspx[^].

See also: http://msdn.microsoft.com/en-us/library/6wd819wh.aspx[^].

Here is a simple tutorial on parsing command line: http://www.cplusplus.com/forum/articles/13355/[^].

—SA

Posted 20-Aug-11 20:05pm

Sergey Alexandrovich Kryukov

Comments

Member 7766180 21-Aug-11 2:18am

Thank you so very much! Yeah it is weird, I can't seem to understand the logic behind it. Here is the first part that appears at the begining, maybe this will help?

if (argc<=2)
{
printf("\nUseage...");
printf("\ndood [IP-address] [packet-count] (ml) (o)");
printf("\n--> eyeball ");
return EXIT_SUCCESS;
}

Sergey Alexandrovich Kryukov 21-Aug-11 2:33am

You're welcome.
If this is not your code you don't have to understand it. Who knows how many stupid things can people write? Make you own code right.
Good luck, call again.
--SA

XTAL256 22-Aug-11 20:56pm

That first part is checking the size of arguments passed to the application.
In that particular example, the application requires the user to provide at least two arguments. If there are less (if (argc<=2)), it will print an explanation of it's usage and state that it needs at least the arguments [IP-address] and [packet-count]. The 3rd and 4th are optional.

Sergey Alexandrovich Kryukov 22-Aug-11 21:06pm

If some arguments are optional, they still have to be processed somewhere; maybe below the shown part of code.
Thank you.
--SA

XTAL256 22-Aug-11 21:36pm

Yes, it looks like those optional args are processed in the first bit of code he posted