Why is argc an 'int' (rather than an 'unsigned int')?

Question

Why is the command line arguments count variable (traditionally argc) an int instead of an unsigned int? Is there a technical reason for t

Answer 1

The fact that the original C language was such that by default any variable or argument was defined as type int, is probably another factor. In other words you could have:

  main(argc, char* argv[]);  /* see remark below... */

rather than

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

Edit: effectively, as Aaron reminded us, the very original syntax would have been something like

  main(argc, argv) char **argv {... } 

Since the "prototypes" were only introduced later. That came roughly after everyone had logged a minimum of at least 10 hours chasing subtle (and not so subtle) type-related bugs