Question

我尝试使用std::cin数组从char[]读取字符。

这是一个简单的程序：

#include <iostream>
#include <conio.h>
#include <iomanip>

using namespace std;

int main() {    
    int age, years;
    char name[20];

    cout << "Enter your age in years: " << endl;

    cin >> years;
    cout << "Enter your name in years: " << name[15] << endl;
    age = years * 12;

    cout << " Your age in months is: " << age << endl;

    cout << "Your name is: " << name[15] << endl;

    return 0;
}

这就是我得到的输出

Enter your age in years:
19
Enter your name in years:
 Your age in months is: 228
Your name is:

它没有从std::cin识别数组。

任何人都可以提供帮助？

Answer 1

#include <iostream>
#include <conio.h>
#include <iomanip>

using namespace std;

int main(){
    int age  , years ;

    char name[20];

    cout <<"Enter your age in years: "<< endl;
    cin >> years;

    cout <<"Enter your name in years: " <<endl;
    cin >> name;
    age = years*12;
    cout << " Your age in months is: " << age <<endl;
    cout << "Your name is: "<< name <<endl;

    return 0;
}

您的代码有两个不同之处：

cin << name;
(...)
cout << name << endl;

我假设你想到

cout＆lt;＆lt; ＆＃34;输入你的姓名＆＃34; ＆LT;＆LT; name [15]＆lt;＆lt; ENDL;

会让它要求输入。这不是cout的作用。 Cout打印出控制台上的内容，它不会要求输入。这是cin的任务。

此外，您不要在数组名称之后放置[15]，这只会打印出数组中的第15个字符，只要输入的名称长度不到15就会成为垃圾字符。

Answer 2

int age, years;
char name[20];
cout << "Enter your age in years: " << endl;
cin >> years;
cout << "Enter your name in years: " << endl;
age = years * 12;
cin >> name;
cout << " Your age in months is: " << age << endl;
cout << "Your name is: " << name << endl;
cin.get();

cin使用char类型数组

2 个答案: