今天在學習《深入理解C++11》中關於移動構造函數的內容時,發現了這麼一個問題:
#include <iostream>
#include <vector>
#include <numeric>
#include <stack>
#include <string>
#include <bitset>
using namespace std;
class HasPtrMem
{
public:
HasPtrMem() :d(new int{0}) {
cout << "Constrcut" << ++n_cstr << endl;
}
HasPtrMem(const HasPtrMem & h) :d(new int(*h.d)) {
cout << "copy Constrcut" << ++n_cptr << endl;
}
/*HasPtrMem(HasPtrMem && h) :d(h.d) {
h.d = nullptr;
cout << "move Constrcut" << ++n_mvtr << endl;
}*/
~HasPtrMem() {
delete d;
cout << "destruct : " << ++n_dstr << endl;
}
int *d;
static int n_cstr;
static int n_dstr;
static int n_cptr;
//static int n_mvtr;
};
int HasPtrMem::n_cstr = 0;
int HasPtrMem::n_dstr = 0;
int HasPtrMem::n_cptr = 0;
//int HasPtrMem::n_mvtr = 0;
HasPtrMem GetTemp() {
//HasPtrMem h;
//cout << "Resource from " << __func__ << ": " << hex << h.d << endl;
//return h;
return HasPtrMem();
}
int main() {
HasPtrMem a = GetTemp();
//cout << "Resource from " << __func__ << ": " << hex << a.d << endl;
system("pause");
return 0;
}
這段代碼在我的Visual studio2015中並不輸出拷貝構造函數的調用,讓我百思不得其解,後來通過蒐集網上資料得知,是因爲編譯器自動優化了臨時對象的傳遞,因此將不會調用copy constructor(生成了默認的移動構造函數 ref:深入理解C++11p83)