C+中的参考行为+； 我可以用StcExpRoad确认C++中引用的内容是否正确。
假设我们有C+中的参考行为+； 我可以用StcExpRoad确认C++中引用的内容是否正确。,c++,reference,C++,Reference,假设我们有 vector<int> a; // add some value in a vector<int> b = a; // 1. this will result another exact copy of inclusive of a's item to be copied in b right? vector<int> &c = a; // 2. c will reference a right? c and a both "point"
vector<int> a;
// add some value in a
vector<int> b = a; // 1. this will result another exact copy of inclusive of a's item to be copied in b right?
vector<int> &c = a; // 2. c will reference a right? c and a both "point"/reference to a copy of vector list right?
vector<int> &d = c; // 3. d will reference c or/and a right? now a, c, d all reference to the same copy of variable
vector<int> e = d; // 4. e will copy a new set of list from d right (or you can say a or c)?
向量a；
//在一个公司里增加一些价值
向量b=a；//1.这将导致在b中复制另一个精确副本，包括a的项目，对吗？
向量&c=a；//2.c将引用一项权利吗？c和a都是“点”/参考向量列表的副本，对吗？
向量&d=c；//3.d参考c或/和a权利？现在，a、c、d都引用同一个变量副本
向量e=d；//4.e将从d右边复制一组新列表（或者你可以说a或c）？
谢谢 没错，
baa/c/dea/c/dint#include <iostream>
int main() {
int a = 7, b = a, &c = a, &d = a, e = d;
std::cout << "a @ " << &a << '\n';
std::cout << "b @ " << &b << '\n';
std::cout << "c @ " << &c << '\n';
std::cout << "d @ " << &d << '\n';
std::cout << "e @ " << &e << '\n';
return 0;
}
你可以看到
acdbebaa/c/dea/c/dint#include <iostream>
int main() {
int a = 7, b = a, &c = a, &d = a, e = d;
std::cout << "a @ " << &a << '\n';
std::cout << "b @ " << &b << '\n';
std::cout << "c @ " << &c << '\n';
std::cout << "d @ " << &d << '\n';
std::cout << "e @ " << &e << '\n';
return 0;
}
acdbecdaeecda如果将元素添加到
eea/c/dcda/c/dcd