C/C++ unsigned integer overflow

Question

i'm reading an article about integer security . here's the link: http://ptgmedia.pearsoncmg.com/images/0321335724/samplechapter/seacord_ch05.pdf

In page 166,there is said:

A computation involving unsigned operands can never overflow,because a result that cannot be represented by the resulting unsigned integer type is reduced modulo to the number that is one greater than the largest value that can be represented by the resulting type.

What does it mean? appreciate for reply.

Answer 1

It means the value "wraps around".

UINT_MAX + 1 == 0 UINT_MAX + 2 == 1 UINT_MAX + 3 == 2 

.. and so on

As the link says, this is like the modulo operator: http://en.wikipedia.org/wiki/Modulo_operation

Answer 2

No overflow?

"Overflow" here means "producing a value that doesn't fit the operand". Because arithmetic modulo is applied, the value always fits the operand, therefore, no overflow.

In other words, before overflow can actually happen, C++ will already have truncated the value.

Modulo?

Taking a value modulo some other value means to apply a division, and taking the remainder.

For example:

0 % 3 = 0  (0 / 3 = 0, remainder 0) 1 % 3 = 1  (1 / 3 = 0, remainder 1)  2 % 3 = 2  (2 / 3 = 0, remainder 2) 3 % 3 = 0  (3 / 3 = 1, remainder 0) 4 % 3 = 1  (4 / 3 = 1, remainder 1) 5 % 3 = 2  (5 / 3 = 1, remainder 2) 6 % 3 = 0  (6 / 3 = 2, remainder 0) ... 

This modulo is applied to results of unsigned-only computations, with the divisor being the maximum value the type can hold. E.g., if the maximum is 2^16=32768, then 32760 + 9 = (32760 + 9) % (32768+1) = 0.