What does int argc, char *argv[] mean?

Question

In many C++ IDE's and compilers, when it generates the main function for you, it looks like this: int main(int argc, char *argv[]) When I code C++ without an IDE, just with a command line compiler, I type: int main() without any parameters. What does this mean, and is it vital to my program?

Answer 1

these are optional command line arguments...

int argc: the number of arguments (tokens) specified on the command line (and passed to main)

char * argv[] : the array of string tokens passed as arguments on the command line to the main

However, argv[0] contains the name of the executable file.

For example...

The program name is myprog.cpp. It is compiled and linked to myprog.exe. Then, at

the command prompt, to run the program...

prompt :> myprog.exe one two three

Command line argument #1 is myprog.exe <--- actually it is the entire full pathname of the program

Command line argument #2 is one

Command line argument #3 is two

Command line argument #4 is three

Here is the source code

============================================

using namespace std;

#include <iostream>

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

{

for (int iLoop=0; iLoop<argc; iLoop++)

{

std::cout << " command line argument # " << (iLoop+1) << " is >" << argv[iLoop] << "<" << endl;

}

}