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

argv and argc are how command line arguments are passed to main() in C and C++.

argc will be the number of strings pointed to by argv. This will (in practice) be 1 plus the number of arguments, as virtually all implementations will prepend the name of the program to the array.

The variables are named argc (argument count) and argv (argument vector) by convention, but they can be given any valid identifier: int main(int num_args, char** arg_strings) is equally valid.

They can also be omitted entirely, yielding int main(), if you do not intend to process command line arguments.

Try the following program:

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

Running it with ./test a1 b2 c3 will output

Have 4 arguments: ./test a1 b2 c3 

Answer 2

argc is the number of arguments being passed into your program from the command line and argv is the array of arguments.

You can loop through the arguments knowing the number of them like:

for(int i = 0; i < argc; i++) {     // argv[i] is the argument at index i }