DEV Community: Alexandru

Stop Using Meters in C++ Graphics: Building a Real-World Material Mass Calculator

Alexandru — Tue, 16 Jun 2026 11:34:33 +0000

When building software for engineers, architects, or machinists, you quickly learn one golden rule: nobody uses meters on the shop floor or construction site. Everything is measured in millimeters (mm).

Today, we will build a practical C++ console application that estimates the mass of various 3D structural shapes based on their material density and dimensions—optimized for real-world blueprints.

📐 The Physics & Math Behind It

The core formula is straightforward:

Mass = Density × Volume

Since density is typically given in kg/m³ and structural dimensions are in mm, our calculated volume will initially be in cubic millimeters (mm³). To get the correct mass in kilograms, we must divide the final volume by 1,000,000,000 (10⁹) to convert mm³ to m³.

Here are the geometric shapes we will support:

Rectangular Prism: V = l × w × h

Solid Cylinder: V = (π × d² × l) / 4

Hollow Tube: V = [π × (D² - d²) × l] / 4

Hexagonal Prism: V = (3 × √3 / 2) × a² × l

#include <iostream>
#include <cmath>

using namespace std;

int main()
{ 
    char solid;
    double mat_density;
    double mass = 0.0; 

    cout << "Enter the solid type: r(rectangular prism), c(cylinder), t(tube), h(hexagonal prism): ";
    cin >> solid;

    while (solid != 'r' && solid != 'c' && solid != 't' && solid != 'h') {
        cout << "Invalid solid type! Enter r, c, t, or h: ";
        cin >> solid;
    }

    cout << "Enter the material density [kg/m3]: ";
    cin >> mat_density;

    switch(solid){
        case 'r': {
            double l, w, h;
            cout << "Enter the length of the prism [mm]: ";
            cin >> l;
            cout << "Enter the width of the prism [mm]: ";
            cin >> h;
            cout << "Enter the height of the prism [mm]: ";
            cin >> w;

            mass = mat_density * (l * w * h) / 1000000000.0; 
            break;
        }
        case 'c': {
            double d, l;
            cout << "Enter the length of the cylinder [mm]: "; 
            cin >> l;
            cout << "Enter the diameter of the base [mm]: ";
            cin >> d;

            mass = mat_density * M_PI * d * d * l / 4.0 / 1000000000.0;
            break;
        }
        case 't': {
            double D, d, l;
            cout << "Enter the length of the tube [mm]: "; 
            cin >> l;
            cout << "Enter the outer diameter [mm]: ";
            cin >> D;
            cout << "Enter the inner diameter [mm]: ";
            cin >> d;

            if (D <= d) {
                cout << "[Error] Outer diameter must be larger than inner diameter!\n";
                return 0;
            }

            mass = mat_density * l * M_PI * (D * D - d * d) / 4.0 / 1000000000.0;
            break;
        }
        case 'h': {
            double a, l;
            cout << "Enter the length of the prism [mm]: "; 
            cin >> l;
            cout << "Enter the side of the base [mm]: ";
            cin >> a;

            mass = mat_density * (3.0 * sqrt(3.0) / 2.0) * a * a * l / 1000000000.0;
            break;
        }
    }

    cout << "The mass of the solid is: " << mass << " kg" << endl;

    return 0;
}

⚡ Key Takeaways for Clean Code
Hardcoded Constants over pow(): Using 1000000000.0 directly instead of pow(10, 9) avoids runtime floating-point overhead, making compile-time math cleaner.

Early Returns: Catching the impossible case where a pipe's inner diameter is larger than its outer diameter prevents negative mass artifacts.

Initialization Safeguards: Forcing mass = 0.0 at the top ensures that if something catastrophic happens, your system won't output garbage data stored in your RAM.

What shape should we add next? Steel I-beams or asymmetric channels? Let me know in the comments!

Building a Simple Circuit Resistance Calculator in C++ (and a logic trap to avoid)

Alexandru — Mon, 15 Jun 2026 08:38:40 +0000

Every programmer has, at some point, fallen into a logical trap. While building a simple console-based electrical circuit simulator in C++, I stumbled upon a classic boolean logic mistake that completely broke my code.

Here is how to build this useful engineering tool and, more importantly, how to avoid the infinite loop trap!

⚡ The Goal: An Equivalent Resistance Calculator

In physics, calculating the total resistance depends on how components are linked:

Series: R_total = R1 + R2 + ... + Rn
Parallel: R_total = (R1 * R2) / (R1 + R2)

We want a program where the user specifies the number of resistors, inputs their values, and chooses the connection mode (s for serial, p for parallel) interactively.

🚨 The Logical Trap: The Double Non-Equality Bug

When validating user input, my first instinct was to create a while loop that kept asking for the link mode if the user typed an invalid character. It looked like this:

// ❌ THE BUGGY CODE
while (link != 's' || link != 'p') {
    cout << "Invalid link mode. Enter 's' or 'p': ";
    cin >> link;
}

Why this creates an Infinite Loop:
The || (OR) operator means the condition is true if either side is true.
If the user types 's', the first check (link != 's') is False. But the second check (link != 'p') is True (because 's' is not 'p').
Since False OR True = True, the loop runs forever, blocking the program even if you type the correct letter!

🛠️ The Final, Bug-Proof Solution
To fix this, we must use the && (AND) operator. The loop should only run if the input is not 's' AND it is also not 'p'.

Here is the complete, working C++ source code:



#include <iostream>

using namespace std;

int main()
{ 
    int n;
    double R, Rc;
    char link;

    cout << "Enter the number of the resistances in the circuit: ";
    cin >> n;
    cout << "Enter the value of the first resistance (ohms): ";
    cin >> R;
    Rc = R;

    for(int i = 2; i <= n; i++){
        cout << "Enter the value of the current resistance (ohms): ";
        cin >> R;
        cout << "Enter the link mode (s-serial/p-parallel): ";
        cin >> link;

        while(link != 's' && link != 'p'){
            cout << "Invalid link mode. Enter the link mode (s-serial/p-parallel): ";
            cin >> link;
        }

        if(link == 's'){
            Rc = Rc + R;
        }
        else {
            Rc = Rc * R / (Rc + R);
        }
    }

    cout << "The equivalent resistance of the circuit is: " << Rc << " ohms" << endl;

    return 0;
}

Beyond Calculators: Building Automated Motion Reports in SMath Studio

Alexandru — Sun, 14 Jun 2026 19:12:40 +0000

💻 The SMath Cheat Sheet: How to Actually Type It

SMath Studio looks like a text editor, but it behaves like a visual math canvas. If you just type normally, you will get errors. Here is the exact shorthand formula to make everything look like the screenshot:

How to insert Units of Measurement (km/hr, hr, m)
Do not just type the text. SMath needs to know it's a physical unit.
- The Secret: Type a single quote ' right before the unit.
- Example for Velocity: Type v:60*'km/hr. SMath will instantly shade the unit in blue, meaning it recognizes it dynamically.
The Difference Between : and =
- The Assignment (:=): Press the colon key : when you want to store a value or formula inside a variable (e.g., v:60).
- The Evaluation (=): Press the standard equals key = only when you want SMath to calculate and show you the final result (e.g., d=).
How to create the if/else block
- Type f(x):
- Look at the right sidebar panel called "Functions" (or press Ctrl + E if it's hidden).
- Click on the if button. It will automatically generate the logical branch structure with the placeholders on your canvas.
- For the "greater than or equal to" sign (≥), simply type >= on your keyboard.
How to spawn the Graph
- Click on an empty space on your grid.
- Press the @ key (Shift + 2 on most keyboards). This is the universal shortcut that instantly creates the 2D plot box.
- Type f(x) in the bottom-left input field and press Enter.

🚀 Math vs. Strings: Two Ways to Manipulate Digits in C++

Alexandru — Sun, 14 Jun 2026 11:03:37 +0000

In programming, there’s rarely just one way to solve a problem. In this article, we’ll take a classic coding challenge—replacing even digits in a number with 1—and tackle it using two completely different paradigms: pure mathematical logic and modern string manipulation.

Along the way, we’ll explore how to handle tricky edge cases (like 0 and negative numbers) and highlight a common, dangerous C++ rookie mistake: accidentally destroying your variables inside a loop. Whether you are looking for raw performance or clean, readable code, this quick guide has got you covered!



#include <iostream>
#include <string>

using namespace std;

int main()
{
    int number;
    int number_copy;
    int nt = 0; // Stores the result for the first method

    cout << "Enter your number: ";
    cin >> number;

    // --- EDGE CASE HANDLING ---
    // Rule 1: We only accept positive numbers for this algorithm
    if (number < 0) {
        cout << "Please enter a positive number!" << endl;
        return 0;
    }

    // Rule 2: Handing zero separately because it's an even number 
    // and mathematical division loops won't process it.
    if (number == 0) {
        cout << "The changed number for the first method is: 1" << endl;
        cout << "The changed number for the second method is: 1" << endl;
        return 0;
    }
    else {
        // CRITICAL STEP: Back up the original number.
        // The while loop below will destroy the 'number' variable (reducing it to 0).
        number_copy = number;

        long long p = 1; // Base 10 multiplier to reconstruct the number in the correct order

        // --- METHOD 1: THE MATHEMATICAL APPROACH ---
        while (number != 0) {
            int l = number % 10; // Extract the last digit

            // Check if the digit is odd
            if (l % 2 != 0) {
                nt = nt + l * p; // Keep the odd digit as it is
            }
            else {
                nt = nt + 1 * p; // Replace the even digit with 1
            }

            p = p * 10;       // Move to the next decimal place (units, tens, hundreds...)
            number = number / 10; // Remove the last digit from the number
        }
    }
    cout << "The changed number for the first method is: " << nt << '\n';

    // --- METHOD 2: THE STRING MANIPULATION APPROACH ---
    // We use 'number_copy' because 'number' is now 0.
    string numberStr = to_string(number_copy); // Convert the integer to a string

    // Loop through each character of the string
    for (int i = 0; i < numberStr.length(); i++) {
        // Check if the current character represents an even digit
        if (numberStr[i] == '0' || numberStr[i] == '2' || numberStr[i] == '4' || numberStr[i] == '6' || numberStr[i] == '8') {
            numberStr[i] = '1'; // Replace the character with '1'
        }
    }

    // Convert the modified string back into a 64-bit integer (long long) to prevent overflow
    long long nt2 = stoll(numberStr);
    cout << "The changed number for the second method is: " << nt2 << endl;

    return 0;
}

C++ Basics: How to Determine a Triangle Type Using IF-ELSE Statements

Alexandru — Sat, 13 Jun 2026 11:11:55 +0000

A classic problem when learning C++ is checking what type of triangle you have based on 3 sides given by the user. Let's see how to write this logic cleanly using if-else structures without nesting too many conditions.



#include <iostream>
using namespace std;

int main() {
    double a, b, c;
    cout << "Enter three sides of the triangle: ";
    cin >> a >> b >> c;

    // Check if the sides can actually form a triangle
    if (a + b > c && a + c > b && b + c > a) {
        if (a == b && b == c) {
            cout << "The triangle is Equilateral" << '\n';
        }
        else if (a == b || b == c || a == c) {
            cout << "The triangle is Isosceles" << '\n';
        }
        else {
            cout << "The triangle is Scalene" << '\n';
        }
    } else {
        cout << "These sides cannot form a valid triangle" << '\n';
    }

    return 0;
}