Question

I’ve just come across someone’s C code that I’m confused as to why it is compiling. There are two points I don’t understand.

1. The function prototype has no parameters compared to the actual function definition.

2. The parameter in the function definition does not have a type.

---

#include <stdio.h>

int func();

int func(param)
{
    return param;
}

int main()
{
    int bla = func(10);    
    printf("%d", bla);
}

Why does this work?
I have tested it in a couple of compilers, and it works fine.

Answer 1

All the other answers are correct, but just for completion

A function is declared in the following manner:

  return-type function-name(parameter-list,...) { body... }

return-type is the variable type that the function returns. This can not be an array type or a function type. If not given, then int
is assumed.

function-name is the name of the function.

parameter-list is the list of parameters that the function takes separated by commas. If no parameters are given, then the function
does not take any and should be defined with an empty set of
parenthesis or with the keyword void. If no variable type is in front
of a variable in the paramater list, then int is assumed. Arrays and
functions are not passed to functions, but are automatically converted
to pointers. If the list is terminated with an ellipsis (,…), then
there is no set number of parameters. Note: the header stdarg.h can be
used to access arguments when using an ellipsis.

And again for the sake of completeness. From C11 specification 6:11:6 (page: 179)

The use of function declarators with empty parentheses (not
prototype-format parameter type declarators) is an obsolescent
feature.