[xdavidliu]

two things wrong with your statement:

- the ++ operator only acts on integer types, not floats or doubles, so there is no limit to speak of here

- the expression "C++" has value equal to C before incrementing, hence the expression "C/C++" is just one for positive C, even when C is small

[bee_rider]

Hey, wait a minute, ++ is defined for floats.

It might be a bad idea to use it in many cases (since there are values for which the result is just rounded back to the original value), but it works!

[bee_rider]

So, luckily my mistakes cancel out -- C/C++ = 1 always, so it must also in the limit. Once we figure out how to define limit.

[Iwan-Zotow]

> the ++ operator only acts on integer types

no, I believe it works on pointer types and enums as well

[bee_rider]

It is also defined for floats.

Using it seems like a bad move though -- for large values it can round back to the input value.

Indeed, the following stupid test program works, although it may heat up your laptop slightly.

    #include <stdio.h>

    void main()
    {
      float C   = 1.0f;
      float Cin = 0.0f;
      int i=0;

      while (Cin != C)
      {
        i++;
        Cin = C;
        C++;
      }
      printf("%i %e\n", i, C/C++);
    }

And, it finally lets us confirm what the mathematicians never could. When does the limit happen? 16777216. No further questions.