On Sat, Sep 3, 2011 at 8:15 AM, Edward Diener <eldiener@tropicsoft.com>wrote:
On 9/1/2011 9:40 PM, Paul Mensonides wrote:
On Thu, 01 Sep 2011 18:23:16 -0400, Edward Diener wrote:
The remark refers in the resolution above refers to "my reading of the
standard" but does not say which standard is being referred to. Is this part of C99 or C++11 ? I have asked about this on the C++ standard NG and am awaiting an answer.
I am surprised by the evaluation of "#elif constant expression" when the corrwesponding #if statament is true because it means that #if - #elif is not equivalent to #if - #else - #if in this particular case, and I am sure many C++ programmers would have expected that the two were indeed equivalent.
It doesn't "evaluate" it; it just parses it. Even in normal code and, in most cases, even with dynamically-typed code, the compiler or interpreter still has to parse the else-if expressions.
So if I have:
#if 1 #else nonC++gobbledygook #endif
The compiler still parses nonC++gobbledygook and issues an error if it is invalid C++ code ?
No, I believe here Paul is referring to a "regular" if/else if expressions, not preprocessor directives. Neither C99 or C++98 (or C+
+11) are particularly clear on this. To me that means that the solution is on the code side rather than requiring a compiler to allow semi- unspecified behavior in the short term and to alter the standard if it is important enough in the long term.
I admit that I have always thought that an #if - #else - #endif path which is not taken can be anything and does not have to be valid C++. You seem to be saying otherwise.
I think Paul is saying that for #if 1 #elif nonC++gobbledygook #endif the preprocessor may still parse the #elif expression, but it won't be evaluated. If you want to avoid it being parsed, do #if 1 #else #if nonC++gobbledygook #endif #endif - Jeff