Question

我将以下内容作为对STL s generate_n的投入。算法

struct GenerateNumber {     
    GenerateNumber () : i(0) {}
    int operator () (void) {        
        return  i++;
    }
private:
    int i;
};

采用该校长的守则的一个例子是:

std::vector <int> v1 (3);
std::vector <int> v2 (3);
GenerateNumber generateNumber;
std::generate_n (v1.begin (), 3, generateNumber);
std::generate_n (v2.begin (), 3, generateNumber);

但是,结果是,<代码>v1和v2均包含{0,1,2}<>>>,而不是v2,以包含。我用一个断点核实,“s Constructionor”只打过一次(我知道,这并不意味着再一次叫作施工人,但我检查了。)



I know I can solve this by making i static, but I don t understand this behavior. How come the value of i is not kept between consecutive calls?

Answer 1

The generator object is copied when passed to generate_n. Try using std::ref, i.e.

std::generate_n(v1.begin(), 3, std::ref(generateNumber));
std::generate_n(v2.begin(), 3, std::ref(generateNumber));

Edit: Note that std::ref is only available in C++11. It was introduced in TR1 as std::tr1::ref, and is also available in boost as boost::ref.

Answer 2


http://en.cppvis.com/w/cpp/algorithm/generate_n" rel=“nofollow”>std:generate_n  依据价值计算,即复制件。 也许你会检查一下:copy Constructionor/em>正在被点名。

如果没有<代码>std:ref,如果你的问题如你所列举,你可以修改你的校长,以提及电话<编码>的栏目:generate_n:

struct GenerateNumber { GenerateNumber (int& i) : struct GenerateNumber { GenerateNumber () : i(0) {} int operator () (void) { return i++; } private: int& i; }; int main() { int counter = 0; std::vector <int> v1 (3); std::vector <int> v2 (3); GenerateNumber generateNumber(counter); std::generate_n (v1.begin (), 3, generateNumber); std::generate_n (v2.begin (), 3, generateNumber); }

友情链接