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int argc, char *argv[]?

 
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int argc, char *argv[]?, What do they mean?!?
Sn0wm4n
post 15 Jul, 2008 - 11:51 AM
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I've been wondering for a while what these meant (the code in the title) but I finally get to a description of it in the book I'm using to learn C++ but it's hard to understand. Can someone tell me what these mean please?

CODE
#include <iostream>
#include <cstdlib>

using namespace std;

int main(INT ARGC, CHAR *ARGV[]){
.........
}
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gabehabe
post 15 Jul, 2008 - 11:57 AM
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They're basically used when you want to be able to call your program from the command line.

In other words, think of those programs that you call from cmd...

You know how they take flags? example:
shutdown -s -t 1
-s and -t are flags, which are passed to the shutdown program. These are the args that main takes.

In other words, when you write a console program, you only really need int main ()

Hope this helps smile.gif
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jeronimo0d0a
post 15 Jul, 2008 - 12:00 PM
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argc is the number of command line arguements passed to the program which includes it's own name. the array argv is a vector containing them.
argv[0] will be the program name
argv[1] will be the first parameter
ie if I type edit somefile
edit, the program name is argv[0] and somefile is argv[1]
you check argc to make sure there actually are arguements
Hope this helps
Jeronimo
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no2pencil
post 15 Jul, 2008 - 01:35 PM
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http://www.dreamincode.net/forums/showtopic30387.htm
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born2c0de
post 15 Jul, 2008 - 10:21 PM
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QUOTE
int main(INT ARGC, CHAR *ARGV[]){

Keep in mind that int and char should be in lowercase.
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nirvanarupali
post 15 Jul, 2008 - 10:41 PM
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In addition of the posts above.

argc - stands for "argument count"
*argv[] - stands for "argument vector"

However you can change that to any variable names you want instead of using argc and argv.
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polymath
post 16 Jul, 2008 - 08:12 AM
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I didn't know that. Interesting...
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kapax
post 16 Jul, 2008 - 12:05 PM
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Pity, I have nothing else to add sad.gif
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