• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 983
  • Last Modified:

Checking Overflow/Underflow on math

How does one check on overflow/underflow for the following C++ types: long, short, float, double, for add, subtract, multiply, divide.  Example:      long L;      L = 2147483647;
      L++;      // Check flag to see if it overflowed.Using assembly language, I can easily accomplish this under Intel/Win32 via:      seto al      mov fOverflow, alAnd for floating point, I can use _clearfp( ).  However, this is Microsoft compiler and Intel CPU specific.  How does one do this in plain C++??Is there a way?More sample code:      long L;
      L = 2147483647;
      short I;
      I = 32767;
      double d;
      d = 1.7976931348623158e+308;
      d = d * d;
      float f;
      f = 16777216.000000;
1 Solution
There is a Microsoft extension (Structured Exception Handling) in Visaul C++ to do it.

__except( GetExceptionCode() == STATUS_INTEGER_OVERFLOW )

You can get more information by seeing the documentation that comes with Visual C++:
Visaul C++ Books/C/C++/Programming Techniques/Structured Exception Handling/.

ShrifAuthor Commented:
I was very clear in stating in my question that I am seeking a non-platform specific answer.  I cannot rely on assembly language, or Visual C++.In addition, so that it is clear, I need the solution to also work for the largest possible type as well.  I say this because an obvious way to check for overflow on an integer is to perform the same operation with a larget size, a long, and compare the results.  This technique will fail for the largest size, double, which also needs to be handled.
There is an extension to ANSI C that provides for signal (SIGFPE), but it's not guarantted to be implemented by every compiler, and the handling of errno is not standardized either.  Anyway, it would cause a jump to your handler and that's not good for C++.

I would recommend an explicit check as follows:

Check = L = 2147483647;
if (Check != --L) //then overflow

ShrifAuthor Commented:
The proposed solution does not work.  If you add 1 to the value, you may overflow -- that's true, but that doesn't mean that subtracting from the result will not return you back to where you started.  It will.  Sorry.
I am not aware of any 'standard' way to do this but I observe the following:-

For any integer data type (char, int, long, etc)

c = a + b; If b is positive c > a unless overflow
           if b is negative c < a unless underflow

c = a - b; if b is positive c < a unless underflow
           if b is negative c > a unless overflow

c = a * b; abs(c) >= abs(a) if b >= 1 unless overflow
           c == 0 if b = 0

c = a / b; cannot produce over or underflow but can produce
           an error if b == 0

You could encapsulate these rules in a new data type.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

Join & Write a Comment

Featured Post

Upgrade your Question Security!

Your question, your audience. Choose who sees your identity—and your question—with question security.

Tackle projects and never again get stuck behind a technical roadblock.
Join Now