counting odd negative integers

[ryanskyowa]

for the life of me i cant figure out why my program wont count negative odd integers. Here is the code I am using:

int main () {
int odd = 0;
int number = 1;
int even = 0;
while (number !=0) {
cout << "Enter a Number : ";
cin << number;
cout << number;
cout << endl;
if ((number != 0) && ((number % 2) == 0)) {
even = (even + 1);
}
if ((number % 2) == 1) {
odd = (odd + 1);
}
}
cout << even;
cout << " even numbers";
cout << endl;
cout << odd;
cout << " odd numbers";
cout << endl;

[abhishekm71]

Hint: When will number be equal to zero? Your loop will never stop.

while (number !=0) {
 cout << "Enter a Number : ";
 cin << number;
 cout << number;
 cout << endl;
 if ((number != 0) && ((number % 2) == 0)) {
  even = (even + 1);
 }
 if ((number % 2) == 1) {
  odd = (odd + 1);
 }
}

[keskiverto]

count odd negative

I see the program to count both odd and even numbers as long as they are not 0. Even and positive were not requested.

[ryanskyowa]

ok so the whole thing will look like this:

int main () {
int odd = 0;
int number = 1;
int even = 0;
int positive = 0;
int negative = 0;
while (number !=0) {
cout << "Enter a Number : ";
cin << number;
cout << number;
cout << endl;
if ((number != 0) && ((number % 2) == 0)) {
even = (even + 1);
}
if ((number % 2) == 1) {
odd = (odd + 1);
}
if (number < 0) {
negative = (negative + 1);
}
if (number > 0) {
positive = (positive + 1);
}
cout << even;
cout << " even numbers";
cout << endl;
cout << odd;
cout << " odd numbers";
cout << endl;
cout << negative;
cout << " negative numbers";
cout << endl;
cout << positive;
cout << " positive numbers";
cout << endl;
return 0;
}