Question

Why am I able to assign a function that returns an int with no parameters specified to a function pointer that returns an int but takes an int argument?

For example when I do:

int f()
{ return 0;}

int (*ptr)(int) = f;

the compiler gives me no warnings

Answer 1

In C, f doesn’t take “no arguments”, but rather “any arguments”*. Say int f(void) to declare “no arguments”.

_{This is different from C++. Notably, C has separate notions of function “declaration” and function “prototype”:}

int f();              /* declaration -- f is still a mystery */
int f(int, double);   /* prototype -- now we know how to call f */
int f(int n, double c) { return -1; }  /* defintion -- now we can link */

*) As I said in the comment, “any arguments” is restricted to types that do not suffer default-promotion (in the same way as default-promotion happens to variadic arguments). That is, float, all flavours of char and of short int, and also ..., are not permissible in the actual function signature.