返回到int main(），而不是上一个函数

Return to int main() instead of to previous function?

本文关键字：上一个 函数 int main 返回      更新时间：2023-10-16

我是编程新手，这是我设计的第一个程序。这是一个计算器，在一个部分中，我要求用户输入他/她是否想输入另一个和，如果他们说是，重复输入x、输入运算符、输入y，并将它们全部相加。我唯一的问题是，如果你重复这个过程不止一次，然后完成。。。函数返回到它自己（上一个调用方），而不是像我希望的那样返回到main。到目前为止，我已经尝试了#pragma once、#define并使用了一个额外的函数，但解决方案仍然失败了。

我觉得这很难解释，所以这是整个程序，你可以看看。我说的是：

    // Simple Calculator v3.cpp : Defines the entry point for the console application.
//
#include "tots.h"
double userInput()
{
    using namespace std;
    //Prompt the user to enter a digit
    cout << "  Please enter a digit:  " << endl;
    cout << "  ";
    //Declare a variable and assign the user's input to that variable
    double input;
    cin >> input;
    //Return the digit the user entered
    return input;
}
int userOperatorInput()
{
    using namespace std;
    //Prompt the user to enter an operator
    cout << "  Please enter the desired mathematical operation:  "
         << "+ = 1; - = 2, * = 3, / = 4  " << endl;
    cout << "  ";
    //Declare a variable and assign the user's input to that variable
    int op;
    cin >> op;
    //Return the operator the user entered;
    return op;
}
int userFinished()
{
    using namespace std;
    /*Ask the user whether he/she has entered the function he/she desires or if they have not
    yet finished*/
    cout << "  Have you finished the function?  "
         << "yes = 1; no = 0" << endl;
    cout << "  ";
    //Declare a variable and assign the user's input to that variable
    int userfinished;
    cin >> userfinished;
    //Return the user's input
    return userfinished;
}
double calculateSolution(double input1, int op, double input2)
{
    //Determine which operator the user entered and calculate the solution accordingly
    if (op == 1)
        return input1 + input2;
    else if (op == 2)
        return input1 - input2;
    else if (op == 3)
        return input1 * input2;
    else if (op == 4)
        return input1 / input2;
    else return -1;
}
double moreCalculus(double solution)
{
    double xtrainput1 = solution;
    int xtraop = userOperatorInput();
    double xtrainput2 = userInput();
    double xtrasolution = calculateSolution(xtrainput1, xtraop, xtrainput2);
    bool xtrafinished = userFinished();
    if (xtrafinished == 0)
        moreCalculus(xtrasolution);
    else
    {
       #ifndef RETURN
       #define RETURN
        return xtrasolution;
       #endif
    }
}
void displaySolution(double input1, int op, double input2, double solution)
{
    using namespace std;
    cout << "  " << input1 << "  " << op << "  " << input2 << "  =  " << solution << endl;
}
void displayXtraSolution(double xtrasolution)
{
    using namespace std;
    cout << setprecision(20);
    cout << "  Solution:  " << xtrasolution;
}
void end()
{
    using namespace std;
    int x;
    cin >> x;
}
int main()
{
    //Input from the user
    double input1 = userInput();
    //Operation user requires
    int op        = userOperatorInput();
    //Second input from user
    double input2 = userInput();
    /*Declare and assign a boolean that informs whether the user has finished
      the operation or not*/
    bool finished = userFinished();
    //Calculate the solution
    double solution = calculateSolution(input1, op, input2);
    //If the user has finished, display the solution
    /*If the user hasn't finished, assign the required inputs from the user and calculate the
      solution*/
    if (finished == 1)
    {
        displaySolution(input1, op, input2, solution);
    }
    else
    {
        double xtrasolution = moreCalculus(solution);
        displayXtraSolution(xtrasolution);
    }
    /*Finally, display some information on screen and ask whether the user requires to calculate
      any more operations or if he/she wants to terminate the application*/

请注意，我制作了一个包含"stdafx.h"和<iostream>的头文件。

您应该关注循环，如果您的代码需要停止，您只需中断它。

循环时使用do

    int main(){
    do{
    //your code section
    cout<<"Do you want to continue(Y/N):";
    cin>>ch;
    }while(ch=='Y'||ch=='y');
    }
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