vector<int> v = {23,12,56,10};v.push_back(5);v.push_back(25);v.pop_back;autoi = v.insert(v.begin +1,7);cout<<"The size of the given vector "<< v.size <<endl;if(v.empty) {cout<<"Vector is empty"<<endl;}else{cout<<"Vector is not empty"<<endl;}cout<<"Element a...
#include<bits/stdc++.h>usingnamespacestd;intmain{vector<int> arr1 = {1,2,3,4};vector<int> arr2 = {};vector<float> arr3 = {1.2,3.8,3.0,2.7,6.6};cout<<"Size of arr1: "<< arr1.size <<endl;cout<<"Size of arr2: "<< arr2.size <<endl;cout<<"Size of arr3: "<< arr...
int fun( sample b[], int n) { int m= b[0].x; for(int i=0; i<n;i++) if(b[i].x<m) m= b[i].x ; return m; } int main() { sample a[6]; int arr[]={12,6,21,7,10,9}; for( int i=0;i<6;i++) a[i].setx(arr[i]); cout<<fun( a, 6 )<<endl; retu...
已知数组arr的定义如下:( ). int arr[5] = {1,2,3,4,5}; 下列语句中输出结果不是2的是 A. cout 〈< *arr+1 〈 B. cout 〈< *(arr+1)〈〈endl; C. cout 〈〈 arr[1] D. cout 〈< *arr 〈〈endl; 相关知识点: 试题来源: 解析 D ...
inta = 1; int*p = &a; 指针运算 inta = 1; int*p = &a; intb = *p; //解引用 intarr[ 5] = { 1, 2, 3, 4, 5}; p = arr; cout<<*p<< endl; p++; //后移一位 cout<<*p<< endl; p--; //前移一位 cout<<*p<< endl; cout<<*(p+ 2)<< endl; //该元素之后第二个...
cout 用于在屏幕上显示消息,应该是 console output 的简写。它是 C++ 中 ostream 对象,该类被封装在 <iostream> 库中,该库定义的名称都放在命名空间...
cout << "flag的值是:" << flag << endl; ``` 这段代码会输出:flag的值是:1(true) 8. 输出数组元素: ``` int arr[5] = {1, 2, 3, 4, 5}; cout << "数组arr的元素依次是:"; for (int i = 0; i < 5; i++) { cout << arr[i] << " "; } cout << endl; ``` 这段...
intmain() {/*char* str = NULL; setmemory(&str, 100); strcpy(str, "hello"); cout << str << endl;*/intarr[] = {6,7,8,9,10};//int* ptr = arr;int* mp =arr;//*(++ptr) += 123; //arr[1] = 130cout << *mp <<""<< *(mp++) <<""<< *mp <<endl; //6 6 ...
cout <<"两组字符串不相等"<< endl; } } cout <<"两组字符串相等"<< endl; }else{ cout <<"两组字符串不相等"<< endl; }return0; }voidtransform(string* str,intlen){for(autoit = str->begin(); it != str->end(); it++)
cout << arr2 << endl; } 然而,当我运行这个命令时,arr输出ints数组的第一个元素的地址(如预期的那样),但arr2不输出chars数组的第一个元素的地址;它实际上输出"ciao"。 我错过了什么,或者我还没有了解到什么? 对于const void*和const char*,运算符<<超载。char数组转换为const char*并传递给该重载,因为...