escaping - How do I pass in the asterisk character '*' in bash as arguments to my C program?

Question

Let's say I have a C program, and I run it from bash:

$ ./a.out 123 *

The program would output all the command line arguments, but it will show these instead:

Argument 1: 123
Argument 2: a.out

What can I do in my program to fix this?

Answer 1

The shell is replacing the asterisk with the name of each file in the directory.

To pass a literal asterisk, you should be able to escape it:

$ ./a.out 123 *