passing more parameters than specified in prototype in C

Question

I'm practicing reading others' code, picked up one of the simplest package in the GNU core-utils, GNU yes. coreutils-8.26 to be specific, because my machine is debian stretch. Code direct links are at the end.

In main(), we have this line:

  parse_long_options (argc, argv, PROGRAM_NAME, PACKAGE_NAME, Version,
                  usage, AUTHORS, (char const *) NULL);

PROGRAM_NAME, PACKAGE_NAME, Version, AUTHORS shall just be some #define macro for constant c-string, thus we indeed passed 8 parameters to parse_long_options here.

In long-options.h which we included, we declared parse_long_options as following, which accepts only 6 arguments:

void parse_long_options (int _argc,
                     char **_argv,
                     const char *_command_name,
                     const char *_package,
                     const char *_version,
                     void (*_usage) (int),
                     /* const char *author1, ...*/ ...);

So, my question is, how does this package, GNU yes, actually compile and run? It's an UB, at least in C99. (1) Is it just because how GCC is implemented, and hence shall be avoided?

code direct links:

yes.c from coreutils-8.26: (2)

long-options.c from coreutils-8.26: (3)

long-options.h from coreutils-8.26: (4)

ftp of GNU: (5)

score 2 · Accepted Answer · answered Dec 25 '18 at 09:29

Notice that ... at the end of the function declaration? That means the function is a varags (variable argument or variadic) function, similar to e.g. printf or scanf.

It means the function can take any number of arguments of any type, beyond the six specified.

Exactly what kind of arguments you can pass should be specified in the documentation, but I guess it's a list of null-terminated byte strings, and the last need to be a NULL pointer to mark the end of the list.

passing more parameters than specified in prototype in C

1 Answers1