Question

I am reading chapter 2 of Advanced Linux Programming:
http://www.advancedlinuxprogramming.com/alp-folder/alp-ch02-writing-good-gnu-linux-software.pdf

In the section 2.1.3 Using getopt_long, there is an example program that goes a bit like this:

int main (int argc, char* argv[]) {
  int next_option;
  // ...
  do {
    next_option = getopt_long (argc, argv, short_options, long_options, NULL);
    switch (next_option) {
       case ‘h’: /* -h or --help */
       // ...
    }
    // ...

The bit that caught my attention is that next_option is declared as an int. The function getopt_long() apparently returns an int representing the short command line argument which is used in the following switch statement. How come that integer can be compared to a character in the switch statement?

Is there an implicit conversion from a char (a single character?) to an int? How is the code above valid? (see full code in linked pdf)

Answer 1

Neither C nor C++ have a type that can store “characters” as values with some dedicated character-specific properties. In that sense, there’s no “character” type neither in C nor in C++.

In both C++ and C languages char is an integral type. It contains numbers. It is just a smallest (in terms of range) integral type. Conversion between char and int exists, just like it exists between int and long or int and short. char has no special status among other integral types (aside from the fact that char type it is distinct from signed char type).

A literal of the form 'h' in C++ has type char, but as any other integral type it is comparable to int. That’s why you can use it in case label the way it is used in your original example.

In other words, your original code is as “strange” as

switch (next_option) {
       case 1L: ...
       // ...
    }

would be. In this case the switch argument is an int, but the case label is a long. The code is valid. Do you find it surprising? Probably not. Your example with 'h' is in not much different.