Obscure C code (was Re: Excessive optimization)

On Fri, 2010-12-03 at 11:32 +0100, Christian Corti wrote: The compiler returns either 0 or 1 for the unary inversion operator. The C standard dictates that 0 is false, anything else is true. Problem is, 'b' can contain damn near anything and still be true. So "FT"[b] with b=0x80 -- b is still true, but you get an access violation, or possibly just a crap result. If your boolean is a true boolean (i.e. only 0 and 1 are valid values, for false and true respectively) then the code *will* work. So you use the unary negation operator, which turns your 0x80 (true) into 0 (false). "TF"[0] is "T", because the original value was true. Similarly, !0 = 1, and "TF"[1] is "F". However: the compiler would still be correct to use the value 123 instead of 1, thus that code may not work... So it's compiler dependent, but most compilers use 0/1 for boolean results, meaning you might not spot the Big Nasty Bug until much, much later. A better way: b?'T':'F' If b is true, then this statement evaluates to 'T', otherwise it evaluates to 'F'. Just it does it without tricky pointer math, and is more likely to work across different compilers. -- Phil. classiccmp at philpem.me.uk http://www.philpem.me.uk/