CODEWITHSUNDEEP
c++c++11initializationoperator-precedence
eonil
eonil

Reputation: 86155

C++ evaluation order between brace-or-equal initializer and initialization-list?

Where I have this struct,

struct
AAA
{
    AAA() : bbb(2)
    {
        // ccc ???
    }

    int bbb = 1;
    int ccc = bbb;
};

AFAIK, if there's an initialization-list :bbb(2), the expression bbb = 1 will be ignored. And then, it's vague to me what ccc will become finally.

Which one of initialization-list or brace-or-equal initializer would be evaluated first? What's the rule between them?

Upvotes: 7

Views: 1141

Answers (2)

user743382
user743382

Reputation:

The rule was always that fields are always initialised in order of declaration, and C++11 didn't change that. That means bbb's initialiser runs first, then ccc's initialiser runs. It doesn't matter whether either initialiser is specified on the field or as part of the constructor.

Upvotes: 11

Joachim Isaksson
Joachim Isaksson

Reputation: 181077

The C++11 draft §12.6.2.9 says;

If a given non-static data member has both a brace-or-equal-initializer and a mem-initializer, the initialization specified by the mem-initializer is performed, and the non-static data member’s brace-or-equal-initializer is ignored.

[ Example: Given

struct A {
  int i = /∗ some integer expression with side effects ∗/ ; 
  A(int arg) : i(arg) { }
  // ...
};

the A(int) constructor will simply initialize i to the value of arg, and the side effects in i’s brace-or- equal-initializer will not take place. — end example ]

Since initialization is done in declaration order (§12.6.2.10) with the addition of this rule, the value of bbb and ccc will both be 2.

Upvotes: 12

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