vector<string> v6 = {"hi","my","name","is","lee"};for(vector<string>::iterator iter = v6.begin(); iter != v6.end(); iter++) { cout<< *iter <<endl;//下面两种方法都都可以检查迭代器是否为空cout << (*iter).empty() <<endl; cout<< iter->empty() <<endl; } 上面是正向...
cout << (*iter).x << " " << (*iter).y << endl; (*iter).y += 100; } //第二种遍历方式,迭代器修改元素值成功 cout << "第二种遍历方式,迭代器访问修改元素值" << endl; for (vector<Point>::iterator iter = m_testPoint.begin(); iter != m_testPoint.end(); iter++) { cout...
vector<int> v(n)表示声明一个容器v,并给他预定存储空间。每一个单元初始化为0,因此,vector<int> v(n)也等同于vector<int> v(n, 0)。当然,如果想要初始化为其他值,也可以改为vector<int> v(n, val) vector<int> v(5); cout << v.size() << endl; for (int i = 0; i < v.size(); ...
#include <vector>using namespace std;int main(){int a[5] = {1,2,3,4,5};vector<int> str_a; //初始化为空vector<int> str_a1(4, 88); // 定义四个元素,每个元素的值为88;vector<int> str_a2 = str_a1; //把a1的值复制给a2;vector<int> str_a3(str_a1.begin(), str_a1.end(...
<vector>//structtypedef struct student{ char* school_name; char gender; int age; bool is_absent;} StudentInfo;typedef std::vector<StudentInfo*> StudentInfoPtrVec;void print(StudentInfoPtrVec*stduentinfoptrvec){ for (int j=0;j<(*stduentinfoptrvec).size();j++) { std::cout<<(*stdu...
}std::cout<<"Printing initial cell center positions ..."<<std::endl;for(unsignedinti =0; i < insideCellCenters.size(); i++) {CVectorcenterPos = insideCellCenters[i]; rawData.initCellCenters.push_back(centerPos);std::cout<<" "; center...
= v.end(); iter+ ) cout << *iter << end 9、l;2. 对于二维vector的定义。1)定义一个10个vector元素,并对每个vector符值1-10。#include<stdio.h>#include<vector>#include <iostream>using namespace std;void main()int i = 0, j = 0;/定义一个二维的动态数组,有10行,每一行是一个用一个...
可以通过以下方式访问vector中的元素: “` vector v = {1, 2, 3, 4, 5}; cout << v[0] << endl; // 输出第一个元素1 cout << v.at(2) << endl; // 输出第3个元素3 cout << v.front() << endl; // 输出第一个元素1
vector::iteratoritePre; cout<<"eraseVECinwrongway"<<endl; for(itePre=myVec.begin();itePre!=myVec.end();itePre++) { myVec.erase(itePre); } printVec(myVec); 按我以前的理解,这样的循环删除方式预期的结果应该是会把vector中的数据清空,但是事实并非如此事实会导致程序崩溃因为itePre迭代器本身...
vector<char*>::iterator it1; char* elem="123"; it1 = find(ve1.begin(), ve1.end(), elem); cout<<*it1; } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. ...