c++ - lambda 捕获变量的规则

Question

例如:

class Example
{
public:
    explicit Example(int n) : num(n) {}
    void addAndPrint(vector<int>& v) const
    {
        for_each(v.begin(), v.end(), [num](int n) { cout << num + n << " "; });
    }
private:
    int num;
};

int main()
{
    vector<int> v = { 0, 1, 2, 3, 4 };

    Example ex(1);
    ex.addAndPrint(v);
    return 0;
}

当您在 MSVC2010 中编译并运行它时，您会收到以下错误:

错误 C3480:“Example::num”:lambda 捕获变量必须来自封闭函数作用域

但是，对于 g++ 4.6.2(预发布版)，您将获得:

1 2 3 4 5

根据标准草案，哪个编译器是正确的？

Answer 1

5.1.2/9:

The reaching scope of a local lambda expression is the set of enclosing scopes up to and including the innermost enclosing function and its parameters.

和 5.1.2/10:

The identifiers in a capture-list are looked up using the usual rules for unqualified name lookup (3.4.1); each such lookup shall find a variable with automatic storage duration declared in the reaching scope of the local lambda expression.

由于num既没有声明在任何函数范围内，也没有自动存储期限，因此无法捕获。因此 VS 是对的，g++ 是错的。