Question

此程序要求您输入三个朋友和他们的披萨片的直径然后它计算比萨饼的面积，看看谁拥有最大的切片。

我需要一些帮助，试图将getline放入我的字符串函数中以摆脱代码中的重复。

#include <iostream>
#include <cmath>
#include <iomanip>
#include <string>
using namespace std;

string presentStringPrompt(string);

double presentDoublePrompt(string);

double computeAreaOfCircle(double);


const double pi = 3.14159;


int main() {


    string first;
    string second;
    string third;
    double diameter;
    double radius;
    double area1;
    double area2;
    double area3;

    cout << "Enter the name of the first friend: \0";
    getline(cin, first);
    cout << "Enter " << first << " pizza diameter : \0";
    cin >> diameter;
    cin.ignore();

    radius = diameter / 2;
    area1 = pi * radius * radius;

    cout << "Enter the name of the second friend: \0";
    getline(cin, second);
    cout << "Enter " << second << " pizza diameter : \0";
    cin >> diameter;
    cin.ignore();

    radius = diameter / 2;
    area2 = pi * radius * radius;

    cout << "Enter the name of the third friend: \0";
    getline(cin, third);
    cout << "Enter " << third << " pizza diameter : \0";
    cin >> diameter;

    radius = diameter / 2;
    area3 = pi * radius * radius;

    cout << showpoint << fixed << setprecision(1);

    cout << " The area of " << first << "'s pizza is " << area1 << " inches squared\0" << endl;
    cout << " The area of " << second << "'s pizza is " << area2 << " inches squared\0" << endl;
    cout << " The area of " << third << "'s pizza is " << area3 << " inches squared\0" << endl;

    if (area1 >= area2 && area1 >= area3) {
        cout << first << " has the largest slice of pizza." << endl;
    }
    else if (area2 >= area1 && area2 >= area3) {
        cout << second << " has the largest slice of pizza." << endl;
    }
    else {
        cout << third << " has the largest slice of pizza" << endl;
    }

    cin.ignore();
    cin.get();
    return 0;
}

string presentStringPrompt(string) {
    string value;

    getline(cin,value);
    return value;
    cin.ignore();
}

double presentDoublePrompt(string) {

}

double computeAreaOfCircle(double) {

}

Answer 1

您之前格式化文本提取的输入流中还剩$ ./bin/cbd C/c : Encode D/d : Decode Q/q : Quit > c ? abcd Binary rep of a= 01100001 (A-65)(S-83)(X-88)(J-74)(D-68)(D-68)(Y-89)(H-72) Binary rep of b= 01100010 (C-67)(O-79)(F-70)(H-72)(B-66)(I-73)(Z-90)(D-68) Binary rep of c= 01100011 (S-83)(U-85)(V-86)(X-88)(P-80)(Q-81)(G-71)(H-72) Binary rep of d= 01100100 (G-71)(K-75)(F-70)(H-72)(U-85)(J-74)(T-84)(Z-90) > d ? 1100001 a [97] > d ? 1100011 c [99] > foo error: invalid choice. > Q个字符。

这就是为什么随后的第一个'\n'语句失败的原因。

您可以在此处获取有关如何解决此问题的更多信息：

Why does std::getline() skip input after a formatted extraction?

Answer 2

在主内部你必须调用这个功能

create or replace function notifications_check_user_id_trigger()
  returns trigger as $$
begin
  if new.notified_user_id <> 
    (select answerer_id from answers where id = new.answer_id)
  then
    raise exception e'answer_id (%) and notified_user_id (%) don\'t match',
      new.answer_id,
      new.notified_user_id;
  end if;
end;
$$ language plpgsql;

create trigger t_50_notifications_check_user_id_trigger
  before insert or update
  on notifications
  for each row execute procedure notifications_check_user_id_trigger()
;

int main()
{
     cout << "Enter the name of the first friend: \0";
     first = presentStringPrompt(first);//or you can use pass by reference that way you don't have to do first = 
     cout << "Enter " << first << " pizza diameter : \0";
     cin >> diameter;
     cin.ignore();
}

Answer 3

模板怎么样？

template <typename T> question(char * question)
{
    T reply;
    cout << question;
    cin >> reply;
    cin.ignore();
    return reply;
}

用作

double diameter = question("Enter " + first + " pizza diameter :");

很少注意到：

如果对字符串使用“”，则不需要终止字符，它由编译器添加。
返回后不添加行，不会执行。

我需要帮助搞清楚如何将getline放入presentStringPrompt以摆脱重复的代码

3 个答案: