#include <iostream>
#include <math.h>
using namespace std;
int sum (int x);
int factorial (int y);
int greatest (int p, int q, int r);
int percentage (int m1, int m2, int m3, int m4, int m5);
int formula (int r, int h);
int voter_age (int x);
int interest (int p, int r, int t);
void swap (int x, int y);
int tables (int i, int j, int k);
int distance (double x, double y, double z);
int speed (double x, double y, double z);
int power (double x, double y, double z);
int sqroot (double x, double y, double z);
int converter (double x, double y, double z);
int inr (double x, double y, double z);
int usd (double x, double y, double z);
int mtrs(double x, double y, double z);
int main () {
int a, b;
double x, y, p, q, m1, m2, m3, m4, m5, r, h, t, i, j, k, z, ans;
cout << "MAIN MENU";
cout << "\n 1. Sum of Natural nos.";
cout << "\n 2. Factorial";
cout << "\n 3. Greatest number among 3";
cout << "\n 4. Percentage(5 subjects)";
cout << "\n 5. Volume of cylinder";
cout << "\n 6. Vote age checker";
cout << "\n 7. interest_calculator";
cout << "\n 8. Swapping nos.";
cout << "\n 9. Table of a number.";
cout << "\n 10. Distance calculator";
cout << "\n 11. Speed calculator";
cout << "\n 12. Calculate the power of a number";
cout << "\n 13. Square root calculator";
cout << "\n 14. Converter";
cout <<
"\n \n Select one of the above option typing the serial number of the
same";
cin >> a;
switch (a) {
case 1:
cout << "Enter any number";
cin >> x;
ans = sum(x);
cout << ans;
break;
case 2:
cout << "Enter any number";
cin >> y;
ans = factorial(y);
cout << ans;
break;
case 3:
cout << "Enter 3 numbers";
cin >> p >> q >> r;
ans = greatest(p, q, r);
cout << ans << " is the biggest number";
break;
case 4:
cout << "Enter marks of 5 subjects ";
cin >> m1 >> m2 >> m3 >> m4 >> m5;
ans = percentage(m1, m2, m3, m4, m5);
cout << ans << "%";
break;
case 5:
cout << "enter value of radius" << "\n";
cin >> r;
cout << "Enter value of height" << "\n";
cin >> h;
ans = formula(r, h);
cout << ans;
break;
case 6:
cout << "Enter your age" << "\n";
cin >> x;
ans = voter_age(x);
break;
case 7:
cout << "Enter principle amount " << "\n";
cin >> p;
cout << "Enter rate " << "\n";
cin >> r;
cout << "Enter time " << "\n";
cin >> t;
ans = interest(p, r, t);
cout << ans;
break;
case 8:
cout << "Enter 1st number.";
cin >> x;
cout << "Enter 2nd number";
cin >> y;
break;
case 9:
cout << "Enter a number to display its table";
cin >> i;
ans = tables(i, j, k);
cout << ans;
break;
case 10:
cout << "Enter speed in km/hr" << endl;
cin >> x;
cout << "Enter time in hours" << endl;
cin >> y;
ans = distance(x, y, z);
cout << ans << "km";
break;
case 11:
cout << "Enter distance in km" << endl;
cin >> x;
cout << "Enter time in hours" << endl;
cin >> y;
ans = speed(x, y, z);
cout << ans << "km/hr.";
break;
case 12:
cout << "Enter a number" << endl;
cin >> x;
cout << "Enter the power" << endl;
cin >> y;
ans = power(x, y, z);
cout << ans;
break;
case 13:
cout << "Enter a number" << endl;
cin >> x;
ans = sqroot(x, y, z);
cout << ans;
break;
case 14:
cout << "Select one of the following" << endl;
cout << "\n a. Currency";
cout << "\n b. Distance";
cout << "\n c. mass";
cout << "\n d. temperature" << endl;
cin >> b;
switch (b) {
case 1:
cout << "Select one of the following:" << endl;
cout << "\t 1. For INR to USD type " << endl;
cout << "\t 2. For USD to INR type " << endl;
cin >> b;
switch (b) {
case 1:
cout << "Enter amount in INR" << endl;
cin >> y;
ans = inr(x, y, z);
cout << ans << "$";
break;
case 2:
cout << "Enter amount in USD" << endl;
cin >> y;
ans = usd(x, y, z);
cout << ans << "Rs.";
break;
}
break;
case 2:
cout << "Slect one of the following" << endl;
cout << "\t Mtrs to kms and cms" << endl;
cout << "\t Kms to Mtrs and cms" << endl;
cout << "\t Cms to Mtrs and Kms" << endl;
break;
case 3:
cout << "Select one of the following" << endl;
cout << "\t Kgs to grams and pounds" << endl;
cout << "\t Grams to Kgs and Pounds" << endl;
cout << "\t Pounds to kgs and grams" << endl;
break;
case 4:
cout << "Select one of the following" << endl;
cout << "\t Celcius to Farenhite and Kelvin" << endl;
cout << "\t Farenhite to Celcius and Kelvin" << endl;
cout << "\t Kelvin to Celcius and Farenhite" << endl;
break;
}
break;
default:
cout << "please enter correct option";
}
}
int sum (int x)
{
int i, sum = 0;
for (i = 1; i <= x; i++)
sum = sum + i;
return (sum);
}
int factorial (int y)
{
int i, fact = 1;
for (i = 1; i <= y; i++)
fact = fact * i;
return (fact);
}
int greatest (int p, int q, int r)
{
int s;
if ((p > q) && (p > r))
s = p;
else if ((q > p) && (q > r))
s = q;
else if ((r > p) && (r > q))
s = r;
return (s);
}
int percentage (int m1, int m2, int m3, int m4, int m5)
{
int s, q;
s = m1 + m2 + m3 + m4 + m5;
q = s / 5;
return (q);
}
int formula (int r, int h)
{
return (r * r * h * 3.14);
}
int voter_age (int x)
{
if (x >= 18)
cout << "eligible to vote";
else if (x < 18)
cout << "Not eligible to vote, wait for " << 18 - x << " years";
return (x);
}
int interest (int p, int r, int t)
{
return (p * r * t) / 100;
}
void swap (int x, int y)
{
x = x + y;
y = x - y;
x = x - y;
cout << "Value of x is " << x << "Value of y is " << y;
}
int tables (int i, int j, int k)
{
for (j = 1; j <= 10; j++)
{
k = i * j;
cout << i << "*" << j << "=" << k << "\n";
}
return (k);
}
int distance (double x, double y, double z)
{
z = x * y;
return (z);
}
int speed (double x, double y, double z)
{
z = x / y;
return (z);
}
int power (double x, double y, double z)
{
z = pow (x, y);
return (z);
}
int sqroot (double x, double y, double z)
{
z = sqrt (x);
return (z);
}
int inr (double x, double y, double z)
{
z = y * 69.70;
return (z);
}
int usd (double x, double y, double z)
{
z = (1 / 69.70) * y;
return (z);
}
int mtrs(double x, double y, double z)
{
z = (1/1000)*y;
return(z);
}
这段代码是我的学校项目，要求我们使用switch创建函数。除了转换器的输出（在第一个开关的情况下）或必须提供十进制输出的任何其他程序的输出之外，其他所有东西都工作正常。 从菜单中选择转换器时，所有的操作都被编程为以小数点形式输出，但是四舍五入。
答案 0 :(得分：2)
请注意，整数类型（char，unsigned int，(u)int<n>_t，size_t）只能容纳整数值。因此，如果为它们分配某种浮点类型的值，则始终会丢失小数。
让我们以distance为例：
int distance (double x, double y, double z)
{
z = x * y; // distance calculated as double!
return (z); // double is cast to int -> you lose the decimals
}
如果要保留小数，请返回浮点类型：
double distance (double x, double y, double z);
// ^^
但是，还有其他一些问题：
首先，不要在返回值上使用括号！！！它们具有特殊含义（创建 reference ），并且可能会给您带来意想不到的结果：
auto distance (double x, double y, double z)
// ^ (!)
{
return (z);
}
在这里，推导返回类型，它将获得对局部变量z的引用，因此您最终将获得未定义的行为！
那为什么要完全将z作为参数传递呢？您永远不会使用它，因此请改为使用局部变量：
double distance (double x, double y)
{
double z = x * y;
return z;
}
或更短，根本不使用中间变量并直接返回（在简短计算中首选此样式）：
return x * y;
有时，您希望具有其他输出参数，然后可以将它们作为参数传递-但是要能够在函数外部接收任何值，则需要将它们作为引用或指针传递。如果必须始终提供值 ，则首选引用，仅当nullptr也被视为有效输入时才使用指针。
int distance (double x, double y, double& z)
// ^ (!)
{
z = x * y;
return z;
}
// use:
double distance;
int rounded = distance(10.12, 12.10, distance);
在此示例中，您有两个结果值，在double变量中计算的距离（带小数）和在int变量中切除的带小数的距离。请注意，将double转换为int时可能会溢出！
以上是一个非常糟糕的示例，由于输出是多余的，如果其中一个输出具有不同/独立的含义，您将执行以下操作：
int distance (double x, double y, double& z)
{
// check input variables x and y
if(...)
{
return INVALID_PARAMETERS; // assuming you have an enum or a #define for
}
// calculations and other checks, different return values for different errors
z = x * y;
return SUCCESS;
}
这有点C编程风格，在C ++中，考虑是否可能抛出一些异常更为合适。具有输出参数的另一种方法是返回结构或类–考虑在结构“点”中返回2D或3D坐标或在结构中具有实部和虚部的复杂结果–好吧，猜想–“复杂”（请注意，不过已经有std::complex。
答案 1 :(得分：0)
除了将返回类型从int更改为double ...
尝试与fixed和setprecision一起运行：
fixed不会删除多余的0，而setprecision会在小数点后的一些小数位数后减少。
例如：
#include <iomanip> //add this include
int main()
{
cout<< fixed;
double x = sum(3);
cout<<setprecision(5)<<x;
return 0;
}
/ 未指定 / setprecision（int n）;
设置小数精度
设置十进制精度，用于格式化输出上的浮点值 操作。
表现为成员精度以n作为参数 作为操纵器插入/提取的流（可以是 在输入流或输出流上插入/提取）。
此操纵器在标头
<iomanip>中声明。