chr.istoph, der Blog

// aufgabe_6.cpp : Definiert den Einstiegspunkt für die Konsolenanwendung.
//
#include "stdafx.h"
#include  

using namespace std; 

int addieren(int a, int b);
int subtrahieren(int a, int b);
int multiplizieren(int a, int b);
int dividieren(int a, int b);

double addieren(double a, double b);
double subtrahieren(double a, double b);
double multiplizieren(double a, double b);
double dividieren(double a, double b);


int main()
{
	char c,d;
	int a,b=0;
	double e, f=0, summe;

	do
	{
		do
		{
			cout << "Minirechner: (i)nteger oder (d)ouble ? ";
			cin >> d;
		}while(d != 'i' && d != 'd');
	
		cout << "Zahl 1 = ? ";
		if(d == 'i')
			cin >> a;
		else
			cin >> e;

		do
		{
			cout << "Rechenart (+ - * /) ? ";
			cin >> c;
		} while (c != 43 && c != 45 && c != 42 && c != 47); 
		
		if(d == 'i')
		{
			do
			{
				cout << "Zahl 2 = ? ";
				cin >> b;
			}while(c == 47 && b == 0);
	
			switch(c)
			{
				case '+':
						summe = addieren(a,b);
					break;
				case '-':
						summe = subtrahieren(a,b);
					break;
				case '*':
					summe = multiplizieren(a,b);
					break;
				case '/':
					summe = dividieren(a,b);
					break;
			}
			cout << a << c << b << "=" << summe;
		}
		else
		{
			do
			{
				cout << "Zahl 2 = ? ";
				cin >> f;
			}while(c == '/' && f == 0);

			switch(c)
			{
				case '+':
						summe = addieren(e,f);
					break;
				case '-':
						summe = subtrahieren(e,f);
					break;
				case '*':
					summe = multiplizieren(e,f);
					break;
				case '/':
					summe = dividieren(e,f);
					break;
			}
			cout << e << c << f << "=" << summe;
		}
		
		cout << "\n\nNoch einmal (j/n)? ";
		cin >> c;
	}
	while(c=='j');

	return 0;
}

int addieren(int a, int b)
{
	return a + b;
}
int subtrahieren(int a, int b)
{
	return a - b;
}

int multiplizieren(int a, int b)
{
	return a * b;
}

int dividieren(int a, int b)
{
	if(b)
		return a / b;
	return 0;
}

double addieren(double a, double b)
{
	return a + b;
}
double subtrahieren(double a, double b)
{
	return a - b;
}

double multiplizieren(double a, double b)
{
	return a * b;
}

double dividieren(double a, double b)
{
	if(b)
		return a / b;
	return 0;
}

Veröffentlicht am: 16.11.2005 von: CHR | publiziert in: C, FH-Aachen