Studying the standard, there was no information on how this would expand. I tried this in visual studio 2008 and it does a = a * (b + c); The standard guarantees that it will always expand to this, and not a = a * b + c? Has this expression always expanded for all standard C ++ versions?
Thank.
Yes, it is guaranteed.
a::operator*=(b + c);
EDIT:
Priority is not indicated in a good table in the standard, there is a note to 5/4saying:
5/4
The priority of the operators is not direct, but it can be obtained from the syntax.
The C ++ Reference table is correct.
=, *=, etc. () (. reference).
=, *=, etc.
, , , . ( , *=).
*=
a *= b + c "" , *= .
a *= b + c
, - .
*= . , a = a * (b + c), , , .
a = a * (b + c)
+ , *=, , b + c, a , *= - .
+
* = - . , , - , . .
, * = , , ", ", .
, , :
, , , "a = (a * b) + c", .
, , "" , , , -, ",",.
googling :
http://www.cppreference.com/wiki/language/operator_precedence
I don’t remember all the details, so it’s good to check if you are not sure between similar operators, but you can do most of this by remembering general principles like this.