c++ - 通过数学引用传递

Question

我开了一家汽水店，我想把所选汽水的成本(2 美元)加到“购物车”总金额中。我正在尝试使用按引用传递，但出现以下错误。

我还尝试在加法函数中将 * 添加到 z 整数，但它会导致另一个错误。如何将传递函数应用于数学问题？

错误:[错误] 从“int*”到“int”的无效转换 [-fpermissive]

cout << "Your total is " << addition (x,&z) << '\n';

主要代码:

// Menu Selection
#include <iostream>
using namespace std;

 void passbyreference (int *z) {
    *z=0;
 }

int addition (int a, int b)
{ 
int z;
  z = a + b;
  return z;


}

void printcost (int z){
    cout << "your total is $" << z << endl;
}
int main()
{
  int x=2, y=2, z=0;

  char selection = ' ';
  string name = "";
  string address = "";


  //Ask user for her/his name

  cout << "Please enter your name: ";
  getline(cin, name);

  //display user name
  cout << "Hello "+ name << endl; 

  cout<< "Please enter your address: ";
  getline(cin, address);

  cout << "Your address is "+ address << endl; 

  do
    {
    cout << "\n\n Welcome to the soda shop, please choose your drink. \n\n" << endl;
    // display menu
    cout<<"My Store";
    cout<<"========" << endl;
    cout<<"1 - $2 Cola" << endl;
    cout<<"2 - $2 Root Beer" << endl;
    cout<<"3 - $2 Sprite" << endl;
    cout<<"X - Exit " <<endl<<endl;

    cout<<"Enter selection: ";
    // read user selection
    cin>>selection;    

    switch(selection)
        {
        case '1':
        cout<< "Cola" << endl;
        cout << "Your total is " << addition (x,&z) << '\n';
            break;
        case '2':
        cout<< "Root Beer" << endl;
        cout << "Your total is " << addition (x,z) << '\n';
            break;
        case '3':
        cout<< "Sprite" << endl;
        cout << "Your total is " << addition (x,z) << '\n';
            break;
        case 'X' :
        case 'x':
        cout<<"Thank you" << endl;
            break;
        // other than 1, 2, 3 and X...
        default : cout<<"Invalid selection. Please try again";
        // no break in the default case
        }
    cout<<endl<<endl;
  } while (selection!= 'X' && selection != 'x');

  return 0;
}

Answer 1

我假设您了解指针工作原理的基础知识。

如果你用 int *z; 初始化一个指针，这意味着 z 将持有一个内存地址。

然后将指针分配给一个变量(定义为 int a = 10; )，您可以编写 z = &a。

发生的事情是 &a 给出该变量的内存地址，然后它被分配给 z 这意味着 *z 将给出你解析了 z 持有的内存地址的值。

你在做什么并不清楚，但我假设你正在将 x 和 z 传递给函数，并且你正在以某种方式尝试更新总值在 z 中。

你需要将int addition (int a, int b)修改为int addition (int a, int *b)让它接受一个指针作为a参数，然后您需要使用参数中的相同 *b 来更新总值。

我写了正确的代码。如果这对您有帮助，请投票。

// Menu Selection
#include <iostream>
using namespace std;

 void passbyreference (int *z) {
    *z=0;
 }

int addition (int a, int *b)
{ 
  *b = *b + a;
  return *b;


}

void printcost (int z){
    cout << "your total is $" << z << endl;
}
int main()
{
  int x=2, y=2, z=0;

  char selection = ' ';
  string name = "";
  string address = "";


  //Ask user for her/his name

  cout << "Please enter your name: ";
  getline(cin, name);

  //display user name
  cout << "Hello "+ name << endl; 

  cout<< "Please enter your address: ";
  getline(cin, address);

  cout << "Your address is "+ address << endl; 

  do
    {
    cout << "\n\n Welcome to the soda shop, please choose your drink. \n\n" << endl;
    // display menu
    cout<<"My Store";
    cout<<"========" << endl;
    cout<<"1 - $2 Cola" << endl;
    cout<<"2 - $2 Root Beer" << endl;
    cout<<"3 - $2 Sprite" << endl;
    cout<<"X - Exit " <<endl<<endl;

    cout<<"Enter selection: ";
    // read user selection
    cin>>selection;    

    switch(selection)
        {
        case '1':
        cout<< "Cola" << endl;
        cout << "Your total is " << addition (x, &z) << '\n';
            break;
        case '2':
        cout<< "Root Beer" << endl;
        cout << "Your total is " << addition (x, &z) << '\n';
            break;
        case '3':
        cout<< "Sprite" << endl;
        cout << "Your total is " << addition (x, &z) << '\n';
            break;
        case 'X' :
        case 'x':
        cout<<"Thank you" << endl;
            break;
        // other than 1, 2, 3 and X...
        default : cout<<"Invalid selection. Please try again";
        // no break in the default case
        }
    cout<<endl<<endl;
  } while (selection!= 'X' && selection != 'x');

  return 0;
}