Curioustab

Home » C# Programming » Operators

Which of the following are NOT Relational operators in C#.NET? >= != Not <= <>=

1, 3
2, 4
3, 5
4, 5
None of these

Correct Answer: 3, 5

← Previous Question Next Question→

More Questions from Operators

What will be the output of the C#.NET code snippet given below? int num = 1, z = 5; if (!(num <= 0)) Console.WriteLine( ++num + z++ + " " + ++z ); else Console.WriteLine( --num + z-- + " " + --z );
Which of the following is NOT a Bitwise operator in C#.NET?
Which of the following is NOT an Assignment operator in C#.NET?
Suppose n is a variable of the type Byte and we wish to put OFF its fourth bit (from right) without disturbing any other bits. Which of the following statements will do this correctly?
Which of the following are the correct ways to increment the value of variable a by 1? ++a++; a += 1; a ++ 1; a = a +1; a = +1;
Suppose n is a variable of the type Byte and we wish, to check whether its fourth bit (from right) is ON or OFF. Which of the following statements will do this correctly?
What will be the output of the C#.NET code snippet given below? byte b1 = 0xF7; byte b2 = 0xAB; byte temp; temp = (byte)(b1 & b2); Console.Write (temp + " "); temp = (byte)(b1^b2); Console.WriteLine(temp);
Which of the following are Logical operators in C#.NET? && || ! Xor %
Which of the following statements are correct about the following code snippet? int a = 10; int b = 20; bool c; c = !(a > b); There is no error in the code snippet. An error will be reported since ! can work only with an int. A value 1 will be assigned to c. A value True will be assigned to c. A value False will be assigned to c.
Which of the following statements are correct about the Bitwise & operator used in C#.NET? The & operator can be used to Invert a bit. The & operator can be used to put ON a bit. The & operator can be used to put OFF a bit. The & operator can be used to check whether a bit is ON. The & operator can be used to check whether a bit is OFF.

Discussion & Comments

No comments yet. Be the first to comment!

Join Discussion