How to simplify this code

[antek83]

I think that there is way to many if statements.


#include <iostream>

using namespace std;


//declaring the constant number and naming it as LUCKY


const int LUCKY = 6;


//declaring the functions for introduction, findhowmanyequal and didtheproducthit


void introduction ();

int findhowmanyequal (int a, int b, int c);

void didtheproducthit (int, int);




int main ()

{

int num1, num2, num3;

int calculate=0;



char answer;



//call for "introduction" function



introduction();



do {

cout << "\nPlease type in any three numbers: " << endl;

cin >> num1; cin >> num2; cin >> num3;

cout << "The three original numbers are: " << num1 << "\t" << num2 << "\t" << num3 << endl;



calculate ++;



// call for "findhowmanyequal" function



findhowmanyequal(num1, num2, num3);



// call for "didtheproducthit" function, while giving it three possible product variables



didtheproducthit(num1, num2);

didtheproducthit(num1, num3);

didtheproducthit(num2, num3);



cout << "\nIf you want to continue please press Y, if not press N" << endl;

cin >> answer;


}



while (answer == 'y'); // if user will press the 'y' key for yes then operation will restart from the main; if any other key then it will stop



cout << "The number of sets entered is: " << calculate << endl;

return 0;


}


void introduction () // this function will be showed only once in the header of the input window


{

cout << "Welcome to assignment number 3! \nThis assignment was written by Anton Kramarenko. \nThe lucky number is: " << LUCKY << endl;

}


int findhowmanyequal (int a, int b, int c) // this function checks if entered numbers are equal to the constant "LUCKY" number


{

int sum=0; // to calculate the total times entered data is equal to "LUCKY"



if (a==LUCKY)

sum++;

if (b==LUCKY)

sum++;

if (c==LUCKY)

sum++;



cout << sum << " of the numbers entered is equal to the lucky number " << LUCKY <<endl; // display the total times the input is equal to "LUCKY" number



return sum;



}


void didtheproducthit (int d, int e) // to check if the product of the entered numbers is equal to the "LUCKY" number


{

if (d*e==LUCKY)



cout << d << " multiplied by " << e << " is equal to the " << LUCKY << endl;

else

cout << d << " multiplied by " << e << " is not equal to " << LUCKY << endl;

}

[doug4]

You could change it to:

if ( (a==LUCKY) || (b==LUCKY) || (c==LUCKY) )
    sum++;

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