C++ 第一个C++程序帮助（计算器）

First C++ program help (calculator)

提问by newbie897

I am a beginner to C++ and pretty much programming altogether (besides a little htmland css).

我是 C++ 的初学者，并且几乎完全编程（除了一点html和css）。

I have decided to start my first project for C++.

我决定开始我的第一个 C++ 项目。

A friend recommended me trying to make a simple calculator so here is my first shot. Any pointers would be great too! Not sure exactly what I am missing, if anything, but the error I am receiving is:

一位朋友建议我尝试制作一个简单的计算器，所以这是我的第一次尝试。任何指针也会很棒！不确定我到底缺少什么，如果有的话，但我收到的错误是：

1>------ Build started: Project: CalculatorFinal, Configuration: Debug Win32 ------
1>  CalculatorFinal.cpp
1>c:\users\ramee\documents\visual studio 2010\projects\calculatorfinal
 \calculatorfinal\calculatorfinal.cpp(32): warning C4102: 'calc' : unreferenced label
1>  CalculatorFinal.vcxproj -> c:\users\ramee\documents\visual studio 2010
\Projects\CalculatorFinal\Debug\CalculatorFinal.exe
========== Build: 1 succeeded, 0 failed, 0 up-to-date, 0 skipped ==========

my code is below (apologize if its not formatted correctly on here. This is my first post :D

我的代码如下（如果这里的格式不正确，请道歉。这是我的第一篇文章：D

  // CalculatorFinal.cpp : Defines the entry point for the console application.
  //

  #include "stdafx.h"       // Including header
  #include <iostream>       // Including ioStream
  using namespace std;      // Namespace

  void calc (double x, double y);
  double result;
  double n1,n2;             // Declaring Variables
  char q,operation;


  int main()                
  { 
   cout<<"Welcome to My Calculator" <<endl; // Outputs welcome message
   cout<<""<<endl;                               // Blank Space
   cout<<"INSTRUCTIONS: Input a mathmatical equation" <<endl; // Outputs instruction            
   cout<<"              EX: 2 + 2" <<endl;       // Outputs instruction
   cout<<""<<endl;                               // Blank Space
   cout<<"Operators:"<<endl;                     // Outputs operation header
   cout<<"For Addition, select '+'"<<endl        // Outputs ADD instruction
   cout<<"For Subtraction, select '-'"<<endl;    // Outputs SUB instruction
   cout<<"For Multiplication, select '*'"<<endl; // Outputs MUL instruction
   cout<<"For Division, select '/'"<<endl;       // Outputs DIV instruction
   cout<<""<<endl;                               // Blank Space
   cout<<"To clear, select 'c'"<<endl;  // Outputs clear instruction
   cout<<"To quit, select 'q'"<<endl;   // Outputs QUIT instruction
   cout<<""<<endl;                                                                  // Blank Space
   cout<<"Input a mathmatical equation"<<endl;                                      // Input instructions
   cin>>n1>>operation>>n2;
   calc:(n1,n2);
   cout<<"The answer is:"<<result<<endl;
   std::cin>>q;             // Input "q" to "quit"
   return 0;}

void calc(double x, double y)                                                       // Operator function
    {            x=n1;
    y=n2;

    switch(operation)                                                           // Operator swtich statement
    {case '+':
        result = x + y;
        break;

    case '-':
        result = x - y;
        break;

    case '*':
        result = x * y;
        break;

    case '/':
        result = x / y;
        break;

    default:
        cout<<"Improper equation. Please input a valid mathmatical equation"<<endl;
        cin>>n1>>operation>>n2;
        calc (n1,n2);
    }

 }

回答by Static Cast

Here is a calculator program I wrote based off of yours which is much nicer:

这是我根据您的程序编写的计算器程序，它要好得多：

#include <iostream>
using namespace std;

//Function prototype
int solve(int, int, char);

int main()
 {
 //Declare variables
 int solution, num1, num2;
 char oper;

 //Output
 cout << "Calculator\n----------\n" << endl;
 cout << "Syntax:\n" << endl;
 cout << "1 + 3\n" << endl;
 cout << "Operators: +, -, *, /\n" << endl;
 cout << "Equation: ";

 //Input
 cin >> num1 >> oper >> num2;

 //Solve and output
 solution = solve(num1, num2, oper);
 cout << "Answer: " << solution << endl;

 //Pause [until enter key] and exit
 cin.ignore(); //Enter key from last cin may be passed, ignore it.
 cin.get();
 return 0;
 }

int solve(int num1, int num2, char oper)
 {
 //Switch oper
 switch(oper)
  {
  case '+':
   return num1 + num2;
  case '-':
   return num1 - num2;
  case '*':
   return num1 * num2;
  case '/':
   return num1 / num2;
  default:
   cout << "\nIncorrect operation!  Try again: ";
   cin >> num1 >> oper >> num2;
   solve(num1, num2, oper);
  }
 }

Here are some things to watch out for, from your last program: 1) Function prototypes do not have function names [i.e void func(int)] 2) Use return values [i.e. return result;] 3) Make sure you have semi-colons.

以下是您上一个程序中需要注意的一些事项：1) 函数原型没有函数名称 [即 void func(int)] 2) 使用返回值 [即返回结果；] 3) 确保您有半冒号。

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[OLD POST: cout<<"For Addition, select '+'"<;* // Outputs ADD instruction

[OLD POST: cout<<"For Addition, select '+'"<;* // 输出 ADD 指令

[No ending semi-colon]

[无结尾分号]

FYI:

供参考：

std::cin>>q; // Input "q" to "quit"

std::cin>>q; // 输入“q”以“退出”

std:: not required here. (using namespace std;)

std:: 此处不需要。（使用命名空间 std;）

(remove colon in calc:(n1,n2);)

（在 calc:(n1,n2); 中删除冒号）

--

——

Your program will work now.]

你的程序现在可以工作了。]

回答by Tushar Jadhav

I wouldn't call this as a C++ program. It is a mistake which almost all amateur C++ programmers make. I would call this as a C style of writing C++ programs. Please don't get me wrong but you need to start thinking in object oriented way so that you can leverage the true power of C++.

我不会将其称为 C++ 程序。这是几乎所有业余 C++ 程序员都会犯的错误。我将其称为编写 C++ 程序的 C 风格。请不要误会我的意思，但您需要开始以面向对象的方式思考，以便您可以利用 C++ 的真正力量。

I would recommend you to make a C++ class called calculator and think on the design of the class for a bit before starting to code. I would keep methods such as Add, Subtract, Divide and so on as public and other methods as private. This would also give you a chance to enhance the calculator class in future like say adding memory support to it so that it remembers the last operation or result. Start thinking in object oriented way, in order to avoid spaghetti code which is difficult to manage later.

我建议您制作一个名为计算器的 C++ 类，并在开始编码之前考虑一下该类的设计。我会将 Add、Subtract、Divide 等方法保留为公共方法，将其他方法保留为私有方法。这也将使您有机会在将来增强计算器类，例如为其添加内存支持，以便它记住上次操作或结果。开始以面向对象的方式思考，以避免后期难以管理的意大利面条式代码。

回答by Henrik

I've commented that I have changed from your original source code. and I've checked this code is working on GCC ( G++ ) compiler

我评论说我已经改变了你的原始源代码。我已经检查过这段代码在 GCC ( G++ ) 编译器上工作

Good luck!

祝你好运！

#include "stdafx.h"       
#include <iostream>       
using namespace std;     

void calc (double _x, double _y); // CHANGED
double result;
double n1,n2;            
double x,y; // CHANGED
char q,operation;

int main()                
  { 
    cout<<"Welcome to My Calculator" <<endl; 
    cout<<""<<endl;
    cout<<"INSTRUCTIONS: Input a mathmatical equation" <<endl;
    cout<<"              EX: 2 + 2" <<endl;  
    cout<<""<<endl;                              
    cout<<"Operators:"<<endl;                 
    cout<<"For Addition, select '+'"<<endl;      
    cout<<"For Subtraction, select '-'"<<endl;    
    cout<<"For Multiplication, select '*'"<<endl; 
    cout<<"For Division, select '/'"<<endl;       
    cout<<""<<endl;                               
    cout<<"To clear, select 'c'"<<endl;  
    cout<<"To quit, select 'q'"<<endl;   
    cout<<""<<endl;                      
    cout<<"Input a mathmatical equation"<<endl;
    cin>>n1>>operation>>n2;
    calc(n1,n2); // CHANGED
    cout<<"The answer is:"<<result<<endl;
    std::cin>>q;      
    return 0;
 }

void calc(double _x, double _y) // CHANGED
    {            
    x=_x; // CHANGED
    y=_y; // CHANGED
    switch(operation)
    {case '+':
        result = x + y;
        break;
    case '-':
        result = x - y;
        break;
    case '*':
        result = x * y;
        break;
    case '/':
        result = x / y;
        break;
    default:
        cout<<"Improper equation. Please input a valid mathmatical equation"<<endl;
        cin>>x>>operation>>y; // CHANGED
        calc (x,y); // CHANGED
    }
}

回答by sayed ali

This is my code,it too long but support more operators

这是我的代码，太长了，但支持更多的操作符

#include "stdafx.h"
#include"iostream"
#include"math.h"
#include"iomanip"
#include <string>
#include <sstream>

using namespace std;
double calc(string mystring);
double calc2(string mystring);
double factoriel(double number);
double root(double num1,double num2);
double dowork(int a,int b,string c);
int main(){
cout<<"***************************************************\n";
cout<<"*                                                 *\n";
cout<<"*                   calculator                    *\n";
cout<<"***************************************************\n\n\n";

string inpstring;
cin >> inpstring;



int length_string=inpstring.length();


double result;
if(abs(calc(inpstring))>abs(calc2(inpstring))){
result=calc(inpstring);
}
else if(abs(calc(inpstring))<=abs(calc2(inpstring))){
result=calc2(inpstring);
}

double s;
     s=3.14;
cout<<"\n"<<"\tresult    :     "<<result<<endl; 


system("pause");
}





double calc(string mystring){
int a=0;//just for switchings
int numberofop=0;
int length_string=mystring.length();

string ops;
string myop;
double param1=0;
double  param2=0;
double  result=0;
string first_inp;
string second_inp;
ops="+-*/^%!R";
    int length_ops=ops.length();

for (int i=0;i<=length_string-1;i++){
    if (i==0){
    if(mystring.substr (0,1)=="-"){
            continue;
        }
    }

    for (int j=0;j<=length_ops-1;j++){

        if (!(mystring.substr (i,1).compare(ops.substr(j,1)))){

            numberofop++;
            if (numberofop==1){
                    myop=ops.substr(j,1);

                    first_inp = mystring.substr (0,i);
                    second_inp = mystring.substr (i+1,length_string-1);

                    stringstream(first_inp) >> param1;
                    stringstream(second_inp) >> param2;


                    if (myop=="+"){
                        a=1;
                        }
                    else if(myop=="-"){
                        a=2;
                    }
                    else if(myop=="*"){
                        a=3;
                    }
                    else if(myop=="/"){
                        a=4;
                    }
                    else if(myop=="^"){
                        a=5;
                    }
                    else if(myop=="%"){
                        a=6;
                    }
                    else if(myop=="!"){
                        a=7;
                    }
                    else if(myop=="R"){
                        a=8;
                    }

            }
        }

    }
 }

switch (a){
            case 1:             
                result=param1+param2;
            break;

            case 2:             
                result=param1-param2;
            break;

            case 3:         
                result=param1*param2;
            break;

            case 4:             
                result=param1/param2;
            break;

            case 5:             
                result= pow(param1,param2);
            break;

            case 6:                     
                result= int(param1)% int(param2);
            break;
            case 7:                     
                result= factoriel(param1);
            break;
            case 8:                     
                result= root(param1,param2);
            break;
}




return result;
}





double factoriel(double a){
    cout<<"enter number      \n";


            double i=a;
            double d=1;
            while(i>1){

            d=d*i;
            i--;
            }
            return d;
}

double root(double num1,double num2){
    double result;
    double reverce;
        reverce=1/num2;
        result=pow(num1,reverce);
return result;
}


double calc2(string mystring){
int a=0;//just for switchings
int numberofop=0;
int length_string=mystring.length();
double pi=3.1415;
double teta;
string ops;
string myop;
double param1=0;
double  param2=0;
double  result=0;
string first_inp;
string second_inp;
ops="logsincostancot";
    int length_ops=ops.length();

for (int i=0;i<=length_string-1;i++){
    if (i==0){
    if(mystring.substr (0,1)=="-"){
            continue;
        }
    }

    for (int j=0;j<=length_ops-1;j++){

        if (!(mystring.substr (i,3).compare(ops.substr(j,3)))){

            numberofop++;
            if (numberofop==1){
                    myop=ops.substr(j,3);


                    second_inp = mystring.substr (i+3,length_string-1);


                    stringstream(second_inp) >> param2;


                    if (myop=="log"){
                        a=1;
                        }
                    else if(myop=="sin"){
                        a=2;
                    }
                    else if(myop=="cos"){
                        a=3;
                    }
                    else if(myop=="tan"){
                        a=4;
                    }
                    else if(myop=="cot"){
                        a=5;
                    }


            }
        }

    }
}

switch (a){
            case 1:             
                result=log(param2);
            break;

            case 2: 
                teta=(double(param2)*pi)/180;
                result=sin(teta);
            break;

            case 3: 
                    teta=(double(param2)*pi)/180;
                result=cos(teta);
            break;

            case 4: 
                teta=(double(param2)*pi)/180;
                result=tanf(teta);
            break;

            case 5:             
                teta=(double(param2)*pi)/180;
                result=1/tanf(teta);
            break;


}

return result;
}
double dowork(int a,int b,string c){
string cut;
cut=c.substr(a,b);
double result;
result=calc(cut);
cout<<"\nresult is    "<<result;

 return result;
}

回答by Nawa

I had coded this simple calculator using c++ so that we can add, subtract, divide or multiple as many numbers as we want. example: 2+3+4-2= and then you will get your answer.

我使用 c++ 编写了这个简单的计算器，以便我们可以根据需要对数字进行加、减、除或乘法。例如：2+3+4-2= 然后你会得到你的答案。

#include <iostream>
#include <conio.h>

using namespace std;

int main()
{
    char c;
    while(true){
        cout << "To solve your math problem you can use this syntex: number(+, -, / or *)number=" << endl;
        cout << "you can use as many number as you want." << endl << endl;

        int n, ans;
        char oper;

        cin >> n;
        ans = n;
        cin >> oper;

        while(oper!='='){

            cin >> n;

            if(oper=='+'){
                ans = ans + n;
            }
            if(oper=='-'){
                ans = ans - n;
            }
            if(oper=='/'){
                ans = ans/n;
            }
            if(oper=='*'){
                ans = ans*n;
            }
            cin >> oper;
        }
        cout << "answer: " << ans << endl << endl;

        cout << "Press esc to exit or press any key to continue." << endl << endl;
        c=getch();
        if(c==27){
            break;
        }
    }
    return 0;
}

回答by Dimos

Here's my take at it, without making a class. Completely new from yours.

这是我的看法，没有上课。全新的。

#include <iostream>
#include <cstdlib>
using namespace std;

double a, b;
char operation;


int main()
{
    cout << "Welcome to my calculator program.\n";
    cout << "To make a calculation simply use the following operators -> (+           - * /).\n";
    cout << "To exit, simply write an operation but replace the operator     with 'q'. (eg. 2q3).\n";

    while (operation != 'q'){
        cin >> a >> operation >> b;
        if(std::cin.fail()){
            cout << "Input not numerical. Exiting...";
            exit(1);
        }

        switch (operation)
        {
        case '+':
            cout << "  = " << a + b << '\n';
            break;
        case '-':
            cout << "  = " << a - b << '\n';
            break;
        case '*':
            cout << "  = " << a * b << '\n';
            break;
        case ':':
        case '/':
            cout << "  = " << a / b << '\n';
        }
    }
    return 0;
}