关于C ++ Noob的简单问题 [英] Simple Question on C++ Noob

问题描述

好的家伙所以我在c ++入门书中练习书店计划。这是一本书，但我不能做这个练习：/





写一个程序，读取同一个交易的几个交易/>
ISBN。写下读取的所有事务的总和。 （这是我必须要做的）



我制作了一个代码，它将两个一起添加但是如何修改它以允许无限制的输入。

Okay guys So I have an exercise to do for a book store program in c++ primer. It is a book but I can't do this exercise :/


Write a program that reads several transactions for the same
ISBN. Write the sum of all the transactions that were read. (This is what I have to do)

I made a code that will add two together but How can I modify it to allow unlimited inputs.

#include <iostream>
#include "Sales_item.h"
int main()
{
Sales_item item1, item2;
std::cin >> item1 >> item2; // read a pair of transactions
std::cout << item1 + item2 << std::endl; // print their sum
return 0;
}

我想读取多个输入并允许我结束语句打印结果。这是我知道的一个菜鸟问题。任何帮助将不胜感激：）

I would like to to read multiple inputs and allow me to end statement to print result. It's a noob question I know. Any help would be appreciated :)

推荐答案

循环！ ...创建一个循环，读取输入数据并将其存储到数组中。如果您不想自己进行数组大小管理，请使用某种容器，例如std :: vector（容器只是内部包含实际数据数组的帮助程序。）

Loop! ...create a loop that reads input data and store it off to an array. If you don't want to do the array size management yourself, use a container of some sort such as a std::vector (containers are just helpers that "contain" the actual data array internally).

您可以使用不确定大小的链接列表或考虑http://www.cplusplus.com/reference/vector/vector/vector/然后您需要做的就是循环直到输入结束命令，添加值到你的向量，在输入结束命令之后，获得向量的迭代器创建一个简单的变量来存储和，在迭代循环时对向量中的所有内容求和。





链接列表的概念是相同的，但是一个允许你使用stl（vector）使用另一个方法链表不必。

You could use a linked list, of indeterminate size or consider http://www.cplusplus.com/reference/vector/vector/vector/ Then all you need to do is loop until end command entered, add values to your vector, after your end command entered, get an iterator to the vector create a simple variable to store the sum, sum everything in your vector while you iterate over the loop.


The concept is the same for a linked list, however one allows you to use the stl (vector) the other approach using a linked list doesn't have to.

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