The #if DEBUG in C#

1. the #if DEBUG in C#
2. Implementation of #if DEBUG in C#

This article will discuss the #if DEBUG statement in the C# programming language.

the #if DEBUG in C#

The #if DEBUG instruction operates in the same manner as an if condition in the program, which means that it will determine whether or not the program is now operating in the debug mode.

If we run the program in the debug mode inside Visual Studio, the condition of the program will immediately become true, and it will execute the code block specified for the true scenario.

A conditional directive that starts with a #if directive must conclude with an explicit #endif directive to function properly.

Implementation of #if DEBUG in C#

Let us take an example to understand the working of #if DEBUG in the C# programming language.

Importing the library is the first step, followed by developing an ifDebugBySaad class and a main() method inside that class to launch the program’s operations.

using System;

class ifDebugBySaad {
  static void Main(string[] args) {}
}

Initialize a variable name of string datatype and assign it a value.

String name = "Saad Aslam";

After that, we will construct a condition that will determine whether or not the application is operating in debug mode and if the answer is true, the if block will be carried out.

#if DEBUG
Console.WriteLine("My name is: " + name);

We use the #else statement if the program is not running in debug mode to print a message to the user. The conclusion of an if the #endif directive indicates a block.

#else
Console.WriteLine("The program is not in debug mode");
#endif

If we want the condition #if DEBUG to always evaluate to true, we need to put the #define DEBUG directive at the very beginning of our code before we declare any libraries.

Full Code:

using System;

class ifDebug {
  static void Main(string[] args) {
    String name = "Saad Aslam";
#if DEBUG
    Console.WriteLine("My name is: " + name);
#else
    Console.WriteLine("The program is not in debug mode");
#endif
  }
}

Output:

My name is: Saad Aslam