C ++传递列表作为函数的参数 [英] C++ pass list as a parameter to a function

问题描述

我正在尝试构建一个非常简单的通讯簿.我创建了一个Contact类，通讯录是一个简单的列表.我正在尝试构建一个允许用户将联系人添加到通讯簿的功能.如果我将代码放在函数之外，则可以正常工作.但是，如果我放进去，那是行不通的.我认为这是一个参考传递与价值传递的问题，我没有认真对待.这是该函数的代码:

I'm trying to build a very simple address book. I created a Contact class and the address book is a simple list. I'm trying to build a function to allow the user to add contacts to the address book. If I take my code outside of the function, it works OK. However, if I put it in, it doesn't work. I believe it's a passing by reference vs passing by value problem which I'm not treating as I should. This is the code for the function:

void add_contact(list<Contact> address_book)
{
     //the local variables to be used to create a new Contact
     string first_name, last_name, tel;

     cout << "Enter the first name of your contact and press enter: ";
     cin >> first_name;
     cout << "Enter the last name of your contact and press enter: ";
     cin >> last_name;
     cout << "Enter the telephone number of your contact and press enter: ";
     cin >> tel;

     address_book.push_back(Contact(first_name, last_name, tel));
}

我没有收到任何错误，但是当我尝试显示所有联系人时，我只能看到原始联系人.

I don't get any errors however when I try to display all contacts, I can see only the original ones.

推荐答案

您正在按值传递address_book，因此将复制您传递的内容，并且当您离开add_contact的范围时，所做的更改是迷路了.

You're passing address_book by value, so a copy of what you pass in is made and when you leave the scope of add_contact your changes are lost.

通过参考传递:

void add_contact(list<Contact>& address_book)

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