When I'm defining some variables like this:
int a = pop(), b = pop(), c = pop();
does C++ give a guarantee that a is going to be initialized first, then b and then c? or is the order not defined?
1 Answer
[dcl.decl]/3 says
-3- Each init-declarator in a declaration is analyzed separately as if it was in a declaration by itself.
Which means your code is treated like:
int a = pop();
int b = pop();
int c = pop();
7 Comments
Paul R
Hmm - I'm not convinced that this implies a predictable order of initialisation...
Andy Prowl
Note 97 provides an explanation, but uses the work "usually". I believe, however, that this only means that there are well-defined exceptions to this rule, but it is not "up to the compiler".
Steve Jessop
@JonathanWakely: so arguably a (minor) defect in the standard, if one has to rely on a footnote for a normative statement of the required behavior. But I think you're correct about what the standard intends to define :-) In such moments I usually say that although footnotes aren't normative, they generally are true.
quetzalcoatl
@JonathanWakely: yes, I've just noticed that. [dcl.init] is exactly the place, and in fact in the note #97 (open-std.org/jtc1/sc22/wg21/docs/papers/2011/n3242.pdf) it is quite clearly stated that the
T a,b,c is a "sequence" and is equivalent to T a;T b;T c;. The word "usually" refers not to exception in the comprehension of a sequence, but to the potential collision of identifiers: T a, T, b, c; is surely not equivalent to T a; T T; T b; T c;. Therefore I agree that it is defined, but damn, I'd like it to be written plainly as a proper point, not a footnote.. |
for (int a = pop(), b = pop(), c = pop(); ...)then the question is more interesting. You could put two of the declarations on the previous line, but now they have a different scope