C++ 将列表作为参数传递给函数
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C++ pass list as a parameter to a function
提问by Adrian
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:
我正在尝试构建一个非常简单的地址簿。我创建了一个 Contact 类，地址簿是一个简单的列表。我正在尝试构建一个函数以允许用户将联系人添加到地址簿。如果我将代码放在函数之外，它就可以正常工作。但是，如果我把它放进去，它就不起作用。我相信这是一个按引用传递与按值传递的问题，我没有像我应该的那样对待。这是函数的代码：
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));
}
I don't get any errors however when I try to display all contacts, I can see only the original ones.
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。
采纳答案by ildjarn
You're passing address_bookby value, so a copy of what you pass in is made and when you leave the scope of add_contactyour changes are lost.
您是address_book按值传递的，因此会复制您传入的内容，并且当您离开add_contact更改范围时会丢失。
Pass by reference instead:
改为通过引用传递：
void add_contact(list<Contact>& address_book)
回答by Krizz
Because you are passing list by value, thus it is copied, and new elements are added to a local copy inside add_contact.
因为您是按值传递列表，因此它被复制，并且新元素被添加到add_contact.
Solution: pass by reference
解决方案：通过引用传递
void add_contact(list<Contact>& address_book).
回答by Kerrek SB
Say void add_contact(list<Contact> & address_book)to pass the address book by reference.
说void add_contact(list<Contact> & address_book)通过引用传递地址簿。
回答by Gran Torino
Pass by reference
通过引用传递
void add_contact(list<Contact>& address_book).
