what risk is When function is called and not use function prototypes

C++ and C compilers behaves differently:

• The C++ compiler doesn't accept a function without prototype (unless it is defined in the same source file, before the calling one), hence, has Hans already pointed out, there is no risk, because you can't even compile the code.

On the other hand

• The C compiler accepts the function (it may generate a warning message, depending on selected warning level) assuming a 'default prototype'.
  Now, if you're lucky, the 'default prototype' will fit with the actual function signature. If you're out of luck then the Coding God's Nemesys will flame you with, at beast, incorrect results.

Answering a follow-up Question about deadlock:

Deadlocks and the problems of the function signature has nothing in common. There are no situations when those problems can be considered in any connection. Hans replied with something like: "this depends on what the function is doing".

Deadlocks are all about threads, when two or more threads wait for each other. In such cases, there are no exception, nothing is corrupting or crashing. Those thread are simply stay in wait state infinitely. You could mean something completely different which you called a deadlock by mistake.

—SA