TITLE: conditionals and precedence

(Newsgroups: comp.std.c++, 17 Feb 99) 


BURSIAN: Achim Bursian <36C93553.97689C0C@erlf.siemens.de>

[snip]

>> I have to maintain some code which contains the bad line
>>   i==0 ? i=1 : i=0;
>> Ok, let us not argue that this is bad, bad style. What the programmer
>> wanted to express seems to be  i ? 0 : 1;
>> But I have this line in the code, and my two compilers produce DIFFERENT
>> output from it:


MCCULLOCH: "Ian McCulloch" <ipm105@rsphy1.anu.edu.au>

> One possibility is that cfront is interpreting the statement as
>
> (i == 0 ? i = 1 : i ) = 0;
>
> That is, the left side of the expression returns either i, or i after
> being assigned the value 1.  The right hand side of the expression 
> sets the left side (ie i) to 0.
> 
> However, according to C++ PL 3rd ed, assignment has a higher precedence
> than?:, so (unless there is a bug in cfront!) that is not what the 
> compiler is doing.


CLAMAGE: Steve Clamage <clamage@Eng.Sun.COM>


In fact, the language definition has changed over time in the
interpretation of such expressions.

Originally, C++ followed the C syntax rule:
    conditional_expression:
 logical_or_expression
 logical_or_expression ? expression : conditional_expression

But C++ now has this syntax rule instead:
    conditional_expression:
 logical_or_expression
 logical_or_expression ? expression : assignment_expression

Also changed is the interpretation of the result. Now, if both
legs of the conditional_expression are lvalues, the result is
also an lvalue. You can now write
 b ? x : y = z
which will be interpreted the same as this version:
 *(b ? &x : &y) = z
Under the old rules, it would have been interpreted as
 b ? x : (y=z)

Evidently your two compilers don't implement the same version of
the rules for conditional expressions.