On Fri, Feb 24, 2012 at 12:31 AM, Vicente J. Botet Escriba < vicente.botet@wanadoo.fr> wrote:
IIIUC I think we agree on the goal but not on how to achieve it. I think verified_int should behave as the builtin types works (as much as possible). Let see some examples
{ short b=3; std::cout <<"-b = "<< -b << std::endl; } { unsigned short a=2; short b=-1; std::cout <<"a/b = "<< a/b << std::endl; } { unsigned short a=2; short b=-1; std::cout <<"a*b = "<< a*b << std::endl; } { unsigned short a=2; short b=-3; std::cout <<"a+b = "<< a+b << std::endl; } { unsigned short a=2; short b=3; std::cout <<"a-b = "<< a-b << std::endl; } { unsigned char a=240; unsigned char b=240; std::cout <<"a+b = "<< a+b << std::endl; }
The results are
-b = -3 a/b = -2 a*b = -2 a+b = -1 a-b = -1 a+b = 480
As you can see the built-in types works as I was suggesting you. The overflow problem occurs when you assign a larger (in range) type to a shorter type or on some operations on the larger (signed/unsigned) integer type.
But maybe, what verified_int is modeling is an integer type that doesn't converts to other verified_int types and checks for overflow. Is this what your library is designed for?
Best,
Vicente
Vicente,
Thank you for your thoughtful input. You are correct in that my verified_int class checks for overflow while preventing any implicit conversion of it's underlying representation. This has certain advantages. Consider a small code refactor which is not expected to change behavior (assume no compiler optimizations): // One math operation per line: verified_uint8_t a = 250; uint8_t b = 10; verified_uint8_t temp = a + b; // This will overflow verified_uint8_t c = temp - b; // We already overflowed above //Now two operations on the same line: verified_uint8_t a = 250; uint8_t b = 10; verified_uint8_t c = a + b - b; // This will also overflow (as currently implemented) However, if the result of a+b can return common_type<A,B>::type which will then subtract b, and then be implicitly converted to verified_uint8_t, the overflow will be avoided. But now that I think more about this, the language will cause an overflow on assignment in the first example, and the language will avoid the overflow in the second example due to the type of the temporary being chosen by the compiler. I will definitely put more thought into this. Excellent feedback. Thank you, Ben Robinson
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