C++ Programming/Exercises/Functions/Pages

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  1. include<iostearm.h>
  2. include<conio.h>
void add();
void mul();
void main()

{

  int a,b;
clrscr();
  cout<<"enter 2 num";
 cin>>a>>b;
add(a,b);
mul(a,b);
gatch();

}

void add (inta,intb);
{
 intc;
c=a+b;
 cout<<c<<endl;
}
 void mul(intx,inty);
{
int c;
 c=x*y;
 cout<<c<<endl;
}

EXERCISE 1[edit]

.Write a program that performs arithmetic division. The program will use two integers, a and b (obtained by the user) and will perform the division a/b, store the result in another integer c and show the result of the division using cout. In a similar way, extend the program to add, subtract, multiply, do modulo and power using integers a and b. Modify your program so that when it starts, it asks the user which type of calculation it should do, then asks for the 2 integers, then runs the user selected calculation and outputs the result in a user friendly formatted manner.

Sample run:

Enter two numbers: 3 12
The sum is 15.
Solution

Solution #1

#include <iostream>
using namespace std;

int AddTwo (int addend1, int addend2)  {
	return addend1 + addend2;
}

int main ()  {
	int number1, number2, sum;
	
	cout << "Enter two integers:\n";
	cin >> number1 >> number2;
	sum = AddTwo(number1, number2);
	cout << "\nThe sum is " << sum << ".";
	
	
}

Solution #2

//by blazzer12
//Input two values. Call a function that returns the sum of the values.

#include<iostream>

using namespace std;

int getSum(int, int);

int main()
{
	int number1, number2, sum;
	
	//get values
	cout<<"Enter two integers to add"<<endl;
	
	cout<<"Number1 :";
	cin>>number1;
	cout<<"Number2 :";
	cin>>number2;
	
	//call getSum() and store result in sum
	sum = getSum(number1, number2);
	
	//print result
	cout<<number1<<" + "<<number2<<" = "<<sum;
}

int getSum(int addend1, int addend2)
{
	return addend1 + addend2;
}

The solution in C.

#include <stdio.h>

int add(int, int);

int main()
{
        int a, b, sum;

        printf("A: ");
        scanf("%d", &a);

        printf("B: ");
        scanf("%d", &b);

        sum = add(a, b);

        printf("The sum of %d and %d is %d.\n", a, b, sum);

        return 0;
}

int add(int a, int b)
{
        return a + b;
}


// Another solution:
# include <iostream>

using namespace std;

int sum (int number1, int number2);

int number1;
int number2;

int main()
{
    
    cout<<"Give me a number amigo: ";
    cin>>number1;
    
    cout<<"Give me another number dude: ";
    cin>>number2;
  
    cout<<"The sum of "<<number1<<" and "<<number2<<" is: "<<sum(number1,number2)<<"."<<endl;
    
    return 0;
    
    
}

int sum (int number1, int number2)
{
    return number1+number2;
}

// by neuroalchemist


EXERCISE 2[edit]

Basically the same as exercise 1, but this time, the function that adds the numbers should be void, and takes a third, pass by reference parameter; then puts the sum in that.

Solution

Solution #1

#include <iostream>
using namespace std;

void AddTwo (int addend1, int addend2, int &sum)  {
	sum = addend1 + addend2;
}

int main ()  {
	int number1, number2, sum;
	
	cout << "Enter two integers:\n";
	cin >> number1 >> number2;
	AddTwo(number1, number2, sum);
	cout << "\nThe sum is " << sum << ".";
	
	return 0;
}

Solution #2

//by blazzer12
//adds two integers using a "pass by reference" type function call.

#include <iostream>
using namespace std;

int addNum(int, int, int&);

int main()
{
	int number1,number2,sum;
	
	//get values;
	cout<<"Enter two integers to add"<<endl;
	
	cout<<"Enter Number 1: ";
	cin>>number1;
	cout<<"Enter Number 2: ";
	cin>>number2;
	
	//call addNum to add the numbers
	addNum(number1, number2, sum);
	
	//print sum
	cout<<number1<<" + "<<number2<<" = "<<sum;
 return 0;
}
int addNum(int addend1, int addend2, int &sum)
{
	sum = addend1 + addend2;
}

Solution #3

#include<iostream>

using namespace std;

void add(int,int);

int sum=0;

int main(){
  cout<<"Please input the first number you wanna add:"<<endl;
  int num1;
  cin>>num1;
  cout<<"Please input the second number you wanna ad"<<endl;
  int num2;
  cin>>num2;
  add(num1,num2);
  cout<<"The sum of these two number:"<<num1<<"&"<<num2<<" is:"<<sum<<endl;
  return 0;
}

void add(int a,int b){
  sum=a+b;
}


EXERCISE 3[edit]

Write a recursive function that finds the #n integer of the Fibonacci sequence. Then build a minimal program to test it. For reference see Wikipedia:Fibonacci number.

For any possible natural number "n", the following applies
   fib(n+2) = fib(n+1) + fib(n)
Also, the following are predefined
   fib(0) = 0
   fib(1) = 1
Solution

Solution #1

#include <iostream>

using namespace std;

unsigned fib(unsigned n);

int main()
{
    // Printing the first 20 Fibonacci sequence values
    for (unsigned i = 0; i < 20; i++){
        cout << "fib(" << i << ") = " << fib(i) << endl;
    }
}

unsigned fib(unsigned n)
{
    if (n < 2)
        return n;

    return fib(n-2) + fib(n-1);
}


EXERCISE 4[edit]

Basically the same as exercise 3, although this time you mustn't use recursion.

Solution

Solution #1

#include <iostream>

using namespace std;

unsigned fib(unsigned n);

int main()
{
    // Printing the first 20 Fibonacci sequence values
    for (unsigned i = 0; i < 20; i++){
        cout << "fib(" << i << ") = " << fib(i) << endl;
    }
}

unsigned fib(unsigned n)
{
    if (n < 2)
        return n;

    unsigned prev1 = 0;
    unsigned prev2 = 1;

    for (unsigned i = 0; i <= n-2; i++){
        unsigned temp = prev1 + prev2;
        // Just doing a rotation of values, since only the last two are needed
        prev1 = prev2;
        prev2 = temp;
    }

    return prev2;
}

For extra exercise, give a big number( like 1000000 ) to both exercise 3 and 4 solutions and compare the execution times. Ponder on the results ;)

EXERCISE 5[edit]

Create a calculator that takes a number, a basic math operator (+,-,*,/,^), and a second number all from user input, and have it print the result of the mathematical operation. The mathematical operations should be wrapped inside of functions.

Solution

Solution #1

Hammad city university

#include<iostream>
using namespace std;
void calculator(int num, int num2, int result);
void calculator(int num,int num2,int result)
{
	char op;
	cout<<"\n     Calculator:-  \nEnter Number: " ;
	cin>>num;
	cout<<"Enter operator +,-,*,/,^  : ";
	cin >>op;
	cout<<"Enter second number: " ;
	cin>>num2;
    if(op=='+')result=num+num2;
	if(op=='-')result=num-num2;
	if(op=='*')result=num*num2;
	if(op=='/')result=num/num2;
	if(op=='^')result=num^num2;
	cout<<"result: "<<result<<"\n";
}
main()
{
	int a,b,c;
calculator(a,b,c);
return 0;
}

}

Solution #2

#include <iostream>
float add(float a, float b)
{
	return a + b;
}
float sub(float a, float b)
{
	return a - b;
}
float mul(float a, float b)
{
	return a * b;
}
float div(float a, float b)
{
	if(b != 0)
	{
		return a / b;
		
	}
	std::cout << "Error: division by zero.\n";
	return 0;
}
float pow(float a, float b)
{
	return pow(a, b);
}
float mod(float a, float b)
{
	return fmod(a, b);
}
void calc()
{
	float a, b;
	char op;
	std::cout << "Enter a #: ";
	std::cin >> a;
	std::cout << "Enter an operator (+, -, *, /, ^, %): "; 
	std::cin >> op;
	std::cout << "Enter a second #: ";
	std::cin >> b;
	switch(op)
	{
		case '+':
			std::cout << "Result: " << add(a, b);
			break;
		case '-':
			std::cout << "Result: " << sub(a, b);
			break;
		case '*':
			std::cout << "Result: " << mul(a, b);
			break;
		case '/':
			std::cout << "Result: " << div(a, b);
			break;
		case '^':
			std::cout << "Result: " << pow(a, b);
			break;
		case '%':
			std::cout << "Result: " << mod(a, b);
			break;
		default:
			std::cout << "Error: operator not valid.\n";
	}
}
int main()
{
	calc();
	return 0;
}

Soln #3

// This program will do simple calculations involving + - * / and ^ //


#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int x,y,sum;

    cout << "Enter first number ";
    cin >> x;
    cout << endl;

    cout << "Enter second number ";
    cin >> y;
    cout << endl;


    cout << 1 << " +" << endl;
    cout << 2 << " -" << endl;
    cout << 3 << " *" << endl;
    cout << 4 << " /" << endl;
    cout << 5 << " ^" << endl << endl;
    cout << "What math would you like to do? ";
    cin >> sum;
    cout << endl;


    switch (sum){
case 1:
    sum = x + y;
    cout << "The answer to your addition is " << sum << endl;
    break;

case 2:
    sum = x - y;
    cout << "The answer to your subtraction is " << sum << endl;
    break;

case 3:
    sum = x * y;
    cout << "The answer to your multiplication is " << sum<< endl;
    break;

case 4:
    sum = x / y;
    cout << "The answer to your division is " << sum << endl;
    break;

case 5:
    sum = pow(x,y);
    cout << "The answer to your power function is " << sum << endl;
    break;

default:
    cout << "You have entered an invalid option " << endl;
    break;
    }



    return 0;
}