write program to display odd and even numbers - c++

Question

im trying to write this code but i couldn't
the q is :
by using for loop, write a program to receive input for any 5 numbers and display the total of even an odd numbers. the output should be as shown below
---------------------------------
Enter any 5 numbers: 0 1 3 2 11
0 is not even number.
total exists even = 1
total exist odd = 3
--------------------------------
and this is what i did:
#include<iostream>
using namespace std;
int main()
{
int i,j=0,c=0;
for(i=0;i<5;i++)
{
cout<<"enter 5 numbers "<<i ;
cin>>i;
}
if(i==0)
{
cout<< "0 is not even number"<<endl;
}
else if(i%2==0)
{j++;}
else if(i%2 !=0)
{c++;}
cout<<"total exists even : "<<j<<endl;
cout<<"total exists ODD : "<<c<<endl;
return 0;
}

Answer 1

Going through your code step by step (notice the changed formatting!):
#include<iostream>
using namespace std; // usually considered bad practice
int main()
{
int i, j=0, c=0;
for(i = 0; i < 5; i++)
{
cout << "enter 5 numbers " << i;
cin >> i; // you are overwriting your loop variable!!!
// how do you think your program will go on if you enter
// e. g. 7 right in the first loop run?
// additionally, you did not check the stream state afterwards
// if user entered something invalid (e. g. S), cin sets the
// fail flag and stops further reading - attemps doing so yield
// 0 (since C++11) or don't modify the variable (before C++11)
}
// this section is outside the loop already!
// so you are only checking the number you read in your loop in the very last run
if(i == 0)
{
cout << "0 is not even number" << endl;
}
else if(i % 2 == 0)
{
j++;
}
// this check is redundant: it is the complement to your previous
// check, so if the first went wrong, the second cannot be false any more
// (compare: you did not check for i != 0 either before doing the modulo check)
else /* if(i % 2 != 0) */
{
c++;
}
cout << "total exists even: " << j << endl;
cout << "total exists odd: " << c << endl;
return 0;
}
Changed code:
#include<iostream>
int main()
{
// several serious coding guide lines mandate: only one variable per line:
unsigned int odd = 0;
unsigned int even = 0;
// I used unsigned int here, negative counts are just meaningless...
// I'm consequent in these matters, but range of (signed) int suffices anyway,
// so you can use either one...
// C++ is not C (prior to C99) - keep scope of variables as local as possible
// (loop counter declared within for header, local variable within body)
for(unsigned int i = 0; i < 5u; i++) // (unsigned? see above)
{
std::cout << "enter 5 numbers (" << i << "): ";
int n; // separate variable!
if(!(std::cin >> n))
{
// some appropriate error handling!!! e. g.:
std::cout << "invalid value entered";
return -1;
}
// this now resides INSIDE the for loop
if(n == 0)
{
cout << "0 is not even number" << endl;
}
else
{
// this is an ALTERNATIVE calculation
n %= 2; // gets either 0 or 1...
odd += n;
even += 1 - n;
// (I personally prefer avoiding conditional branches but you *can*,
// of course, stay with the if/else you had before, too...
// - just don't check the complement as shown above)
}
}
cout << "total exists even: " << even << endl;
cout << "total exists odd: " << odd << endl;
return 0;
}
About the unsigned: Sometimes these are of advantage:
void f(int n) { /* need to check for both 0 <= n && n <= max! */ }
void f(unsigned int n) { /* n <= max suffices */ }
but sometimes one has to handle them with care:
for(unsigned int n = 7; n >= 0; --n) { /* ... */ } // endless loop!!!
for(unsigned int n = 7; n-- >= 0;) { /* ... */ } // correct variant
(the first one would have worked with signed int, but it is not the fault of the unsigned type, but the programmer's fault who did not chose the right type for what he or she intended...).
Just for completeness: Assuming we could drop the mathically incorrect statement that zero wasn't even, we could have it even much simpler:
unsigned int constexpr LoopRuns = 5u;
int main()
{
unsigned int odd = 0; // just one single variable...
for(unsigned int i = 0; i < LoopRuns; i++)
{
std::cout << "enter 5 numbers (" << i << "): ";
int n;
if(!(std::cin >> n))
{ /* ... */ }
odd += n %= 2;
}
// one single difference instead of five additions...
cout << "total exists even: " << LoopRuns - odd << endl;
cout << "total exists odd: " << odd << endl;
return 0;
}

Answer 2

This program will help you out.
#include <iostream>
int main () {
int num[5], even = 0, odd = 0;
bool hasZero = false;
std::cout << "Enter 5 numbers:"
for (int i = 0; i < 5; i++) {
std::cin >> num[i];
}
for (int i = 0; i < 5; i++) {
if (num[i] == 0) { // Checking if the current number is zero
hasZero = true;
} else if (num[i] % 2 == 0 ) { // Checking if the current number is even
++even;
} else { // If the number is not even, then it must be odd
++odd;
}
}
if (hasZero) { // If the input has zero then print following statement
std::cout << "0 is not an even number" << std::endl;
}
std::cout << "Total even count: " << even << std::endl;
std::cout << "Total odd count: " << odd << std::endl;
return 0;
}
If you are unable to understand any line, then you're most welcome in the comments section below ;)
The problem with your code:
In the for statement, you're using the same variable for both counter and input , i.e., i. This will allow neither for loop to execute properly nor the input to be captured properly.
You're overwriting the i variable everytime you take any input, then only the last input (out of 5 inputs) will be stored in memory.
You're just checking the last input, by using if statement, because the loop is already ended before.
If you want your code to run properly, then these modifications will make that work:
#include<iostream>
using namespace std;
int main()
{
int num,j=0,c=0; // Change the name to num here, because i will be used later as a counter variable.
for(int i=0;i<5;i++)
{
cout<<"enter 5 numbers "<<i ;
cin>>num;
// Don't end for loop here, this will not allow every input to be checked.
if(num==0)
{
cout<< "0 is not even number"<<endl;
}
else if(num%2==0)
{
j++;
}
else if(num%2 !=0) // Or just add a *else* here instead of *else if*, they will work exactly the same here.
{
c++;
}
} // End of for loop
cout<<"total exists even : "<<j<<endl;
cout<<"total exists ODD : "<<c<<endl;
return 0;
}

Answer 3

Firstly, 0 is an even number, and your code needs to be properly indented, just so you can see that you are indeed reading the input into a single integer, which also controls the loop, and your if statement is outside the for loop (despite the misleading indentation. Here's a simple example implementation, but you can (and should) fix the bugs I pointed out in your own code:
#include <iostream>
int main() {
std::cout << "Enter 5 numbers\n";
int cnt(5);
int n, odd(0), even(0);
while(cnt-- && (std::cin >> n))
n % 2 ? ++odd : ++even;
std::cout << odd << " odd, "
<< even << " even numbers" << std::endl;
return 0;
}
Note the post decrement and the fact the && short-circuits.

Answer 4

This should be your code:
you take an array of integer where you store the input value. Head over to https://www.tutorialspoint.com/cprogramming/c_arrays.htm to learn more abour arrays..
#include<iostream>
using namespace std;
int main(){
int i,j=0,c=0;
int numbers[5];
for(i=0;i<5;i++){
cout<<"enter 5 numbers "<<i ;
cin>>numbers[i];
}
for(i=0;i<5;++i){
if(numbers[i]==0)
{
cout<< "0 is not even number"<<endl;
}
else if(numbers[i]%2==0)
{j++;}
else if(numbers[i]%2 !=0)
{c++;}
}
cout<<"total exists even : "<<j<<endl;
cout<<"total exists ODD : "<<c<<endl;
return 0;
}

Answer 5

using namespace std;
int main()
{
int * Array = new int[5];
int even(0), odd(0);
for(int i = 0; i < 5; i++)
{
cout<<"enter "<< i+1 << "-th number: " << flush;
cin>>Array[i];
if(!Array[i])
{
cout<< "0 is not even number... input again"<<endl;
i = i-1;
}
else
{
if(Array[i]&1) odd++;
else even++;
}
}
cout<<"total exists even : "<<even<<endl;
cout<<"total exists ODD : "<<odd<<endl;
cin.get(); cin.get();
delete[] Array;
return 0;
}