Question

我使用struct和pointers编写了一个代码，问题是当你输入age时，结果将为0。如果你看看我的代码会很有帮助

Person's name: Billy
    Person's age: 25
    Write your hobby[0]:Kungfu 
    Write your hobby[1]:soccer
    write your hobby[2]:basketball
    Write your crush name: Jake
    Name:Billy
    Age:0
    Hobbies[0]:Kungfu
    Hobbies[1]:soccer
    Hobbies[2]:basketball
    Crush: Jake

由于某种原因，Age被分配到25但结果显示0为什么会发生

#include <iostream>
#include <string>
#include <Windows.h>
#include <sstream>
using namespace std;

//declare structure to store info about Billy
struct Son{
    string name;
    string crush;
    int age;
    string hobbies[3];
}Person;

int main(){
    string sAge;
    int i;
    Son* info = new Son;
    info = &Person;
    //user interface
    //Person's name
    cout << "Person's name: ";
    getline(cin, info ->name); //inputs person's name
    //Person's age
    cout << "Person's age: ";
     //inputs person's age
    getline(cin,sAge);
    (stringstream)sAge << info ->age; 
    //for loop to get hobbies
    for(i = 0; i < 3; i++){
        cout << "Write your hobby[" << i <<"]: ";
        getline(cin,info ->hobbies[i]); //inputs the person hobby three times
    }
    //Person's crush
    cout << "Write your crush name: ";
    getline(cin, info ->crush); //inputs the person's crush *opitional*

    //output statement
    cout << "Name: " << info ->name << endl; //display name
    cout << "Age: " << info ->age << endl; //display age
    for(int j = 0; j < 3; j++){ //display hobbies
    cout << "Hobbies[" << j << "]: " << info ->hobbies[j] << endl;
    }
    cout << "Crush: " << info ->crush << endl; //display crush                                             
    delete info;
    system("pause");
    return 0;
}

Answer 1

(stringstream)sAge << info ->age;

这不是正确的方法，您可以使用其中任何一个：

stringstream ss;
ss << sAge;
ss >> info->age;

info->age = atoi(sAge.c_str());

或者，您可以使用std::atoi(sAge)立即接受std::string，但是，由于您有using namespace std;，因此应自动使用string。感谢@alexolut说明这一点。

您基本上将stringstream投射到stringstream这是一个巨大的失败，string不是static_cast的子类。你也不应该在C ++中使用C风格的演员表，学会使用reinterpret_cast，dynamic_cast和Windows.h来产生更好的警告（感谢@JohnZwinck说明这一点）。< / p>

附注：你包括{{1}}并且你没有使用它的任何功能。

C ++年龄不会显示（struct）

1 个答案: