CODEWITHSUNDEEP
c#
Hiddenshadow1234
Hiddenshadow1234

Reputation: 23

Adding doubles appending instead of summing

I am working on a calculator and whenever my calculator adds up a number it just joins them together.

An example of this is 3+5 it gives me 35. I converted the numbers into doubles and it still gives me 35. Here is my code:

Console.Write("Enter a Number: ");
double num1 = Convert.ToDouble(Console.ReadLine());

Console.Write("Enter a Operator: ");
string op = Console.ReadLine();

Console.Write("Enter a Number: ");
double num2 = Convert.ToDouble(Console.ReadLine());

if (op == "+")
{
    Console.WriteLine(num1 + " plus " + num2 + " is " + num1 + num2);
}

Console.ReadLine();

Thank you for viewing my post have a nice day.

Upvotes: 2

Views: 91

Answers (2)

user2819245
user2819245

Reputation:

Because there are strings involved in your expression using +, and because of the order in which those string arguments appear in that expression, the C# compiler treats the + operators as string concatenation.

In more detail: The + are interpreted as string concatenation because the expression is evaluated from left to right. The first evaluated term is num1 + " plus ". Since there is a string involved, the first + is treated as operator for string concatenation. With the first term evaluated to a string, the second and following + each with their respective second argument are then also treated as string concatenation, because the left-hand side of each (partial) term is always being evaluated to a string.

A "cheap" way to fix the code is to use parantheses to help the compiler to distinguish between the arithmetic operator and the string concatenation operator:

Console.WriteLine(num1 + " plus " + num2 + " is " + ( num1 + num2 ) );

A safer and more readable way to write the output can be achieved by using string interpolation (note the $ symbol in front of the string), which conveniently gets rid of the + operator confusion:

Console.WriteLine($"{num1} plus {num2} is {num1 + num2}");

(Thanks to @AlexeiLevenkov for pointing out the latter)

Upvotes: 3

TheGeneral
TheGeneral

Reputation: 81493

The other answers are sufficient and explain the problem, though you might like to use String Interpolation, and double.TryParse to validate the input and give feedback on failure

String Interpolation

The $ special character identifies a string literal as an interpolated string. An interpolated string is a string literal that might contain interpolated expressions. When an interpolated string is resolved to a result string, items with interpolated expressions are replaced by the string representations of the expression results. This feature is available in C# 6 and later versions of the language.

Double.TryParse Method

Converts the string representation of a number to its double-precision floating-point number equivalent. A return value indicates whether the conversion succeeded or failed.

Example

// valid operators
var ops = new[] { "+", "-", "/", "*" };

double num1;
Console.Write("Enter a Number: ");
while (!double.TryParse(Console.ReadLine(), out num1))
   Console.Write("Invalid number, try again");

string op;
Console.Write($"Enter operator (\"+\", \"-\", \"/\", \"*\"): ");
while (!ops.Contains(op = Console.ReadLine()))
   Console.Write("Invalid operator, try again");

double num2;
Console.Write("Enter a Number: ");
while (!double.TryParse(Console.ReadLine(), out num2))
   Console.Write("Invalid number, try again");

// local function to do stuff
double ApplyOperator()
{
   switch (op)
   {
      case "+": return num1 + num2;
      case "-": return num1 - num2;
      case "/": return num1 / num2;
      case "*": return num1 * num2;
      default: throw new InvalidOperationException();
   }
}

// an interpolated string
Console.WriteLine($"{num1} {op} {num2} = {ApplyOperator()}");

Console.ReadLine();

Output

Enter a Number: 1
Enter operator ("+", "-", "/", "*"): +
Enter a Number: 2
1 + 2 = 3

Upvotes: 0

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