Question

@sarnold确定这里是我的更新试图按照你给我的指南我尝试删除主要移动原型到类型，只有每个我缩小到一个int main包含每个单独的函数的声明和对每个函数使用空洞而不是int，希望这是正确的举动。

#include <iostream>
#include <cmath>
#include <stdio.h>
#include <string>

using namespace std;


void function1()
{
    //Enter Variables
    int number;
    cout << "Enter Variable a: ";
    cin >> number;
    int a = number;


    cout << "Enter Variable b: ";
    cin >> number;
    int b = number;
    cout << "Enter Variable c: ";
    cin >> number;
    int c = number;
    if (a > b && a > c);
    return a;
    cout << "This is your highest number(" << a << ")";
    if (b > a && b > c);
    return b;
    cout << "This is your highest number(" << b << ")";
    if (c > a && c > B);
    return c;
    cout << "This is your highest number(" << c << ")";

}

void function2()
{
    // Prompt User for Integer

    int number;
    cout << "Enter integer 1: ";
    cin >> number;
    int a = number;

    cout << "Enter integer 2: ";
    cin >> number;
    int b = number;

    if (a != B)
        cout << "Continue";

    else
        cout << "Number 1 or 2 is the same!";

    cout << "Enter integer 3: ";
    cin >> number;
    int c = number;

    if ((a != B) != c)
        cout << "Continue";
    else
        cout << "Number 3 is the same as number 1 or 2!";
    cout << "Enter integer 4: ";
    cin >> number;
    int d = number;

    if (((a != B) != c) != d)
        cout << "numbers are all different";
    else
        cout << "Number 3 is the same as number 1, 2, or 3!";
    cout << "Thank you enter another number?\n\n";
    cin >> number;
    return 0;
}

std::string numerals = "VXLCDM";
void function3()
{
    char roman_Numeral;
    int arabic_Numeral = 0;
    cout << "Enter the Roman Numeral in Capital letters (e.g. CCXIX) : ";
    while(cin.get(roman_Numeral))
    {
        if(roman_Numeral == 'M')
            arabic_Numeral = arabic_Numeral + 1000;
        else if(roman_Numeral == 'D')
        {
            roman_Numeral = cin.peek();
            if(numerals.find(roman_Numeral, 5) != std::string::npos)
            {
                arabic_Numeral = arabic_Numeral - 500;
                continue;
            }
            else
            {
                arabic_Numeral = arabic_Numeral + 500;
                continue;
            }
        }
        else if(roman_Numeral == 'C')
        {
            roman_Numeral = cin.peek();
            if(numerals.find(roman_Numeral, 4) != std::string::npos)
            {
                arabic_Numeral = arabic_Numeral - 100;
                continue;
            }
            else
            {
                arabic_Numeral = arabic_Numeral + 100;
                continue;
            }
        }
        else if(roman_Numeral == 'L')
        {
            roman_Numeral = cin.peek();
            if(numerals.find(roman_Numeral, 3) != std::string::npos)
            {
                arabic_Numeral = arabic_Numeral - 50;
                continue;
            }
            else
            {
                arabic_Numeral = arabic_Numeral + 50;
                continue;
            }
        }
        else if(roman_Numeral == 'X')
        {
            roman_Numeral = cin.peek();
            if(numerals.find(roman_Numeral, 2) != std::string::npos)
            {
                arabic_Numeral = arabic_Numeral - 10;
                continue;
            }
            else
            {
                arabic_Numeral = arabic_Numeral + 10;
                continue;
            }
        }
        else if(roman_Numeral == 'V')
        {
            roman_Numeral = cin.peek();
            if(numerals.find(roman_Numeral, 1) != std::string::npos)
            {
                arabic_Numeral = arabic_Numeral - 5;
                continue;
            }
            else
            {
                arabic_Numeral = arabic_Numeral + 5;
                continue;
            }
        }
        else if(roman_Numeral == 'I')
        {
            roman_Numeral = cin.peek();
            if(numerals.find(roman_Numeral) != std::string::npos)
            {
                arabic_Numeral = arabic_Numeral - 1;
                continue;
            }
            else
            {
                arabic_Numeral = arabic_Numeral + 1;
                continue;
            }
        }
        else
            break;
    }
    cout << arabic_Numeral << endl;
    return 0;
}
void input (double &);
void convert (double radius, int & feet, double & total);
void output (int, double, double);
void function4()
{
    double total, radius;
    int feet;
    char yn;
    do
    {
        input (radius);
        convert (radius, feet, total);
        output (feet, total, radius);
        cout << "Enter a Different radius? (y or n)\n";
        cin >> yn;
        cout << "\n\n";
    }
    while (yn == 'y'  || yn == 'Y');
    system("pause");
    return 0;
}
void input(double & radius)
{
    cout << "---------------------------Re Enter Radius-------------------------\n\n";
    cout << "Enter Radius:\n";
    cin >> radius;
    cout << "\n";
}
void convert(double radius, int & feet, double & total)
{
    //=======================================================
    total = 3.14 * radius * radius * radius * 4 / 3;
}
//========================================================
void output(int feet, double total, double radius)
{
    cout << "---------------------------Re Enter Radius-------------------------\n\n";
    cout << "The volume is: " << total << " \n\n";
    cout << "---------------------------   Equation    -------------------------\n\n";
    cout << "The Equation with (" << radius << ") being:\n\n";
    cout << "4/3 * 3.14 * (" << radius << ")^3 = " << total << "\n\n";
    cout << "---------------------------     Again?    -------------------------\n\n";
}
void swap( int &x, int &y )
    int number;

void function5()

{

    cout << "x is ";
    cin >> number;
    int x = number;

    cout << "y is ";
    cin >> number;
    int y = number;
    swap(x, y);
    cout << "Swapped the first number is now(" << x << ")and the second is now (" << y >> ")";

    return 0;
}
int leap (int year);
void function6 (void){
    int month,day,year,dm,dn,leap;

    printf("enter the month:");
    scanf("%d",&month);

    printf("enter the day:");
    scanf("%d",&day);

    printf("enter the year:");
    scanf("%d",&year);

    if (leap==0)
    {   if(month==1)
        dm=0;
        if(month==2)
            dm=31;
        if(month==3)
            dm=59;
        if(month==4)
            dm=90;
        if(month==5)
            dm=120;
        if(month==6)
            dm=151;
        if(month==7)
            dm=181;
        if(month==8)
            dm=212;
        if(month==9)
            dm=243;
        if(month==10)
            dm=273;
        if(month==11)
            dm=304;
        if(month==12)
            dm=334;}

    else
    {    if(month==1)
        dm=0;
        if(month==2)
            dm=31;

        if(month==3)
            dm=60;
        if(month==4)
            dm=91;
        if(month==5)
            dm=121;
        if(month==6)
            dm=152;
        if(month==7)
            dm=182;
        if(month==8)
            dm=213;
        if(month==9)
            dm=244;
        if(month==10)
            dm=274;
        if(month==11)
            dm=304;
        if(month==12)
            dm=335;}

        dn=dm+day;

        printf("the day number is :%d",dn);

        return 0;
}

int leap (int year){
    if((year%100== 0 && year%400==0)||  (year%4==0))

        return 1;
    else
        return 0;}

int main(int argc, char *argv[], int year, int leap, int number, int arabic_Numeral)

{
    function1();
    function2();
    function3();
    function4();
    function5();
    function6();
    return 0;
}

Answer 1

好的，你有int main()定义了三次。你需要给每个人一个不同的名字。

此外，您不需要复制#include和using语句，只需在开头使用一次。

我建议将原型移动到课程的开头，只是为了便于阅读。

Answer 2

if (a > b && a > c);
return a;
cout << "This is your highest number(" << a << ")";

这三行总结了代码中的许多问题：

您的if条件无用：
```
if (foo);
```
仅执行foo测试;如果foo为真，则分号括起要执行的代码的 end ：
```
if (foo)
    /* then code */ ;
```
您的return a语句会阻止执行其余代码，包括紧随其后的cout行。

如果您使用文本编辑器（例如vim）可以为您执行语法识别缩进，那么这些小问题中的一些会更加明显。编辑器并不总是像编译器一样好，但是如果你注意它们如何布局你的代码，它们可以帮你节省一些时间。

此外，至少应该由编译器向您报告代码 - { - 1}}。如果没有，请打开警告。（在return上，gcc很不错。）

<强>更新

要将每个文件放入一个文件中并依次调用它们，您需要稍微重新格式化它们。每个程序目前都是自包含的，几乎完全在-Wall -Wextra例程中运行。这在玩具程序中很好，但使您在单个可执行文件中调用它们的工作变得复杂。

目前，在伪代码中你有几个这样的文件：

prog 1：

main()

prog 2：

main() {
    /* ask for input */
    /* run the routine */
    /* maybe call functions */
    /* output results */
    /* quit */
}

您需要像这样重新安排您的计划：

程序：

main() {
    /* ask for input */
    /* run the routine */
    /* maybe call functions */
    /* output results */
    /* quit */
}

在这种模板下，你几乎可以将东西复制粘贴在一起;实际上，您可能希望这样做，尽管全局变量会使问题复杂化。（没有充分理由阻止全球存储的另一个原因。）

如果您有足够的时间在项目到期之前，我强烈建议您稍微退一步并重新编写代码 - 您的例程当前正在混合输入，计算和输出，并且代码通常更容易测试和如果从计算中拆分输入和输出，则保持不变。如果您提供一些帮助功能，您的输入和输出会更容易。快速勾勒出这可能是什么样子：

void run_routine_one(); /* I think C++ forbids a (void) prototype here */
void run_routine_two();

main() {
    run_routine_one();
    run_routine_two();
    /* quit */
}

void run_routine_one() {
    /* ask for input */
    /* run the routine */
    /* maybe call functions */
    /* output results */
}

void run_routine_two() {
    /* ask for input */
    /* run the routine */
    /* maybe call functions */
    /* output results */
}

我希望您能看到将程序分解为更小的部分并使每个功能尽可能少地实现的易读性。你可能会遇到一些大功能，它们确实能够保持大型功能，但却试图将功能分解为尽可能小的功能。任务越小，越有可能发现一些公平简单的重写可以导致更简单的代码。（查看int prompt_for_int(string) { int ret; cout << string; cin >> ret; return ret; } int days_in_month(int month, int year) { int leap = is_leap_year(year); /* J F M A M J J A S O N D */ int days[2][12] = { {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}}; if (month < 0 || month > 11 || year < 1753) return -1; return days[leap][month]; } void program_one() { int month, year, days; month = prompt_for_int("Enter month, 1 for January, 2 for February...\n"); month -= 1; /* zero-indexed month */ year = prompt_for_int("Enter year, not before 1754...\n"); days = days_in_month(month, year); cout << "There are " << days << " in " << month + 1 << " " << year << nl; /* I'm no C++ expert -- perhaps >> nl isn't idiomatic or even functional */ } int main(int argc, char *argv[]) { program_one(); program_two(); /* maybe given them better names */ return 0; }例程 - 它使用数组来存储天数，并根据days_in_month()函数的结果选择要使用的数组。我学会了这个来自The Elements of Programming Style（page 54） - 尽管是为FORTRAN和PL / 1编写的，但它仍然是一本很好的书。无论如何，这个小例程非常易于测试使用一个小程序，因为它本身不做任何输入和输出。）

多功能

2 个答案: