Why does a function with no parameters (compared to the actual function definition) compile?

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.First, the function prototype has no parameters compared to the actual function definition. Second, 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

It's an old compiler glitch;

if the function takes no parameters, then it must be declared void inside parenthesis.

otherwise, the compiler gets the parameter list at declaration time...

Perhaps there is a compiler switch you can turn on and off to eliminate

this deprecated feature