Usage of C Command-line Arguments via Echo Program

tl;dr

When main is called, it is called with two arguments. The first (conventionally called argc, for argument count) is the number of command-line arguments the program was invoked with; the second (argv, for argument vector) is a pointer to an array of character strings that contain the arguments, one per string.

Below is an example code of echo program.

#include <stdio.h>

int main(int argc, char *argv[]) {
  while (--argc > 0)
    printf((argc > 1) ? "%s " : "%s", *(++argv));
  printf("\n");
  return 0;
}

argc and argv

argc is an integer representing the number of arguments, and argv is a pointer to an char array, i.e., a string array, which stores the argument strings. argv contains at least one string, which is the command itself. For example, from the echo command

$ echo hello, world

argv would contain items like {"echo", "hello,", "world"}. Therefore argc >= 1, and argc == 1 implies the program gets no arguments.

Also note that argv has additional null pointer such that argv[argc] == 0.

Pointer to a string array, argv

Since argv is a pointer, assuming it contains items {"echo", "hello,", "world"}, the following statements are valid.

• *argv == argv[0] == "echo"
• *++argv == *(++argv) == argv[1] == "hello,"
• (*argv++)[0] == argv[1][0] == 'e', after the previous increment
• *argv == argv[2] == "world", after the previous increment
• *++argv == argv[3] == argv[argc] == 0 // or null pointer, after the previous increment

References

• The C Programming Language 2E, Kernighan & Ritchie