C++ `argc` and `argv` arguments

Question

Can anyone explain what are and how to use "argc" and "argv"?

It seems my new assignment is about "argc" and "argv", but I don't know what they are.

--

thanks guys but i still dont understand what's what. these are codes from my revision notes, anyone can explain which is which that has been explained below?

class Tank{

public;

int wheels;

};

int main(int argc, char* argv[1]){

Tank sherman;

sherman.wheels =12;

cout << "tank has " << sherman.sheels << "wheels" >> endl;

return 0;

}

Answer 1

These allow you access to the command line arguments that may have been specified while launching your application.

int main (int argc, char** argv)
{
    char* fileName = argv[0]; // The name of your executable

    char* firstArgument = argv[1]; // The first specified command line argument
    char* secondArgument = argv[2]; // The second specified command line argument

    int argumentCount = argc - 1;

    return 0;
}

Answer 2

Try doing some reading: http://www.site.uottawa.ca/~lucia/courses/2131-05/labs/Lab3/CommandLineArguments.html

Answer 3

When you invoke a program, you can supply command line arguments to it. E.g.:

prog-name arg1 arg2

So if you write a program and want to access command line arguments sent to it, you use argc and argv.

In the previous example, arg1 and arg2 are command line arguments. So,

argc = 3
argv[0] = "prog-name"
argv[1] = "arg1"
argv[2] = "arg2"

Real life example:

rm -r doc/

Here,

argc = 3
argv[0] = "rm"
argv[1] = "-r"
argv[2] = "doc/"

Hope that helps!

Answer 4

in C++, argc is the argument count, with the first argument always being the name of the program. argv is an array of character arrays that contain the command line arguments (argv[0] being the name of the program).

one way to use them is like so:

int main(int argc, char** argv)
{
  for(int i = 0; i < argc; ++i)
  {
    std::cout << "argv[" << i << "] = " << argv[i] << std::endl; 
  }

  return 0;
}

Which prints out the command line arguments and exits.

Answer 5

When you execute your program you can give it a number of additional arguments, e.g.:

./myProgram dog cat

dog and cat will be passed into your main function. also the executable itself (./myProgram) is passed as the very first argument.

Because you don't know at the time of writing the code how many arguments you will get, your only chance is to get the arguments in a vector (array) with their count.

void main (int argc, char** argv)

argc (argument count) will hold the number of arguments including the executable name, and argv (argument vector) points to each of them. In this example:

argc = 3
argv[0] = "./myProgram"
argv[1] = "dog"
argv[2] = "cat"

Answer 6

"argc" and "argv" are the arguments that get passed to the entry point "main()". Argument "argc" specifies the number of command line arguments (argument count) and "argv" is an array of strings that contain the actual arguments passed via command line.
For example, consider a program "testprog.exe" has an entry point like this:

int main( int argc, char** argv )
{
    for( int i = 0; i < argc; i++ )
    {
        printf( "\n argument[%d] is %s", i, argv[i] );

    }
}

If the above code is executed as:

testprog.exe hello world

Now argc would be 3, and:

• argv[0] would contain executable name "testprog.exe"
• argv[1] would contain "hello"
• argv[1] would contain "world"

See here for more information:
http://publications.gbdirect.co.uk/c%5Fbook/chapter10/arguments%5Fto%5Fmain.html