ImperialViolet

The title of this post is clearly daft; signed numbers are of a finite size so, of course they overflow. However, physical reality doesn't agree with the C standard which says that compilers can (and do) assume that overflow never happens. Take this, for example:

int a, b;
if (a > 0 && b > 0 && a + b > 0) foo();

A compiler can remove the third test because it's redundant given the assumptions that a + b cannot overflow.

Clearly, this is pretty scary stuff and it's one of the reasons that I use unsigned everywhere. However, I'm very happy to read the GCC 4.2 change log to see the following:

New command-line options -fstrict-overflow and -Wstrict-overflow have been added... With -fstrict-overflow, the compiler may assume that signed overflow will not occur, and transform this into an infinite loop. -fstrict-overflow is turned on by default at -O2, and may be disabled via -fno-strict-overflow. The -Wstrict-overflow option may be used to warn about cases where the compiler assumes that signed overflow will not occur. It takes five different levels: -Wstrict-overflow=1 to 5. See the documentation for details. -Wstrict-overflow=1 is enabled by -Wall.