C++ Program to Check Positive, Negative or Zero

Use a simple if-else structure in C++ to check if a number is positive, negative, or zero, for example: if (num > 0) cout << "Positive"; else if (num < 0) cout << "Negative"; else cout << "Zero";.

Examples

Input5

OutputPositive

Input-3

OutputNegative

Input0

OutputZero

How to Think About It

To decide if a number is positive, negative, or zero, compare it with zero using greater than and less than checks. If it is greater than zero, it is positive; if less, it is negative; otherwise, it is zero.

Algorithm

1

Get the input number from the user.

2

Check if the number is greater than zero.

3

If yes, print 'Positive'.

4

Else, check if the number is less than zero.

5

If yes, print 'Negative'.

6

Otherwise, print 'Zero'.

Code

cpp

#include <iostream>
using namespace std;

int main() {
    int num;
    cout << "Enter a number: ";
    cin >> num;

    if (num > 0) {
        cout << "Positive" << endl;
    } else if (num < 0) {
        cout << "Negative" << endl;
    } else {
        cout << "Zero" << endl;
    }
    return 0;
}

Output

Enter a number: 5
Positive

Dry Run

Let's trace input 5 through the code

1

Input

User enters num = 5

2

Check if num > 0

5 > 0 is true

3

Print result

Print 'Positive'

Step	Condition	Result
Check num > 0	5 > 0	true
Check num < 0	5 < 0	false
Print output	-	Positive

Why This Works

Step 1: Input number

We first get the number from the user using cin.

Step 2: Check positivity

If the number is greater than zero using if (num > 0), it is positive.

Step 3: Check negativity

If not positive, check if it is less than zero with else if (num < 0) to identify negative numbers.

Step 4: Zero case

If neither positive nor negative, the number must be zero, so print 'Zero'.

Alternative Approaches

Using switch with sign function

cpp

#include <iostream>
using namespace std;

int sign(int n) {
    return (n > 0) - (n < 0);
}

int main() {
    int num;
    cout << "Enter a number: ";
    cin >> num;
    switch(sign(num)) {
        case 1: cout << "Positive" << endl; break;
        case -1: cout << "Negative" << endl; break;
        default: cout << "Zero" << endl;
    }
    return 0;
}

This uses a helper function and switch-case for clarity but is slightly more complex.

Using ternary operator

cpp

#include <iostream>
using namespace std;

int main() {
    int num;
    cout << "Enter a number: ";
    cin >> num;
    cout << (num > 0 ? "Positive" : (num < 0 ? "Negative" : "Zero")) << endl;
    return 0;
}

This is a compact one-line decision using nested ternary operators, good for short code.

Complexity: O(1) time, O(1) space

Time Complexity

The program performs a fixed number of comparisons regardless of input size, so it runs in constant time O(1).

Space Complexity

Only a few variables are used, so the space complexity is constant O(1).

Which Approach is Fastest?

All approaches run in O(1) time; the if-else method is simplest and most readable for beginners.

Approach	Time	Space	Best For
If-else	O(1)	O(1)	Simplicity and clarity
Switch with sign function	O(1)	O(1)	Structured code with switch-case
Ternary operator	O(1)	O(1)	Compact code, less readable for beginners