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This section of our 1000+ C# multiple choice questions and answers focuses on Floating and Decimal Data Types in C# Programming Language.

1. Select a convenient declaration and initialization of a floating point number:
a) float somevariable = 12.502D
b) float somevariable = (Double) 12.502D
c) float somevariable = (float) 12.502D
d) float somevariable = (Decimal)12.502D

Answer: c
Explanation: We cannot implicitly convert a “double” number directly to any other data type. Here, its float we have to add the required data type to number as :

float somevariable = (float)12.502D;
 or
Double somevariable = (Double)12.502D;

2. Number of digits upto which precision value of float data type is valid?
a) Upto 6 digit
b) Upto 8 digit
c) Upto 9 digit
d) Upto 7 digit

3. Valid Size of float data type is?
a) 10 Bytes
b) 6 Bytes
c) 4 Bytes
d) 8 Bytes

4. Correct way to define a value 6.28 in a variable ‘pi’ where value cannot be modified?
a) #define pi 6.28F
b) pi = 6.28F
c) const float pi = 6.28F
d)

Answer: c
Explanation: Const is a reserve keyword whenever they are declared with any variables. The value stored in that variable always remain fixed. Any modification done to change value of that variable results in error. Hence, options #define pi 6.28F, pi = 6.28F and const float pi, pi = 6.28F are rejected because the value is not declared fixed. Now, for value only option ‘const float pi = 6.28F’ is correct because for value while declaring a constant variable we need to provide a constant value too as provided in const float pi = 6.28F. So, option const float pi = 6.28F is correct way of declaration of constant variable.

5. What will be the correct set of C# code to display the value of given variable ‘c’ as ‘25.302’.
a)

```
   float a = (double) 12.502f;
```
```
   float b = 12.80f;
```
```
   float c;
```
```
   c = (float) a + b;
```
```
   Console.Writeline(c);
```
```
   Console.ReadLine();
```

```
   float a = 12.502D;
```
```
   float b = 12.80f;
```
```
   float c;
```
```
   c = a + b;
```
```
   Console.WriteLine(c);
```
```
   Console.ReadLine();
```

```
   double a = 12.502;
```
```
   float b = 12.802f;
```
```
   float c;
```
```
   c = (float)a + b;
```
```
   Console.WriteLine(c);
```
```
   Console.ReadLine();
```

```
   double a = (float) 12.502f;
```
```
   float  b = 12.80f;
```
```
   float  c;
```
```
   c = a + b;
```
```
   Console.WriteLine(c);
```
```
   Console.ReadLine();
```

Answer: c
Explanation: Expression

c = (float) a + b.

6. Minimum and Maximum range of values supported by ‘float’ data type are?
a) 1.5 * 10-40 to 3.4 * 1038
b) 1.5 * 10-45 to 3.4 * 1030
c) 1.5 * 10-45 to 3.4 * 1038
d) 1.5 * 10-45 to 3.4 * 1037

7. Select appropriate difference between decimal, float and double data type in C#?

8. Why does a float variable stop incrementing at number ‘16777216’ in the following C# code?

```
 float a = 0 ;
```
```
 while (true)
```
```
 {
```
```
     a++;
```
```
     if (a > 16777216)
```
```
     break;
```
```
 }
```

a) Sign and Exponent for ‘16777217’ is same as for ‘16777216’
b) Mantissa is different for ‘16777216’ and ‘16777217’
c) Sign and Exponent for ‘16777217’ is different from ‘16777216’
d) None of the mentioned

Answer: b
Explanation: 16777216 is exactly 224, and would be represented as 32-bit float like so:
sign = 0 (positive number)
exponent = 24 (stored as 24 + 127 = 151 = 10010111)
mantissa = . 0
As 32 bits floating-point representation: 0 10010111 00000000000000000000000
Therefore: Value = (+ 1) * 2 ^ 24 * (1. 0 + . 0) = 2 ^ 24 = 16777216
Now let’s look at the number 16777217, or exactly 224 + 1:
sign and exponent are the same.
Mantissa should have to be exactly 2-24 so that (+ 1) * 2 ^ 24 * (1. 0 + 2 ^-24) = 2 ^ 24 + 1 = 16777217 and here lies the actual problem. The mantissa cannot have the value 2-24 because it only has 23 bits, so the number 16777217 just cannot be represented with the accuracy of 32-bit floating points numbers.

```
static void Main(string[] args)
```
```
{
```
```
    int x = 1;
```
```
    float y = 2. 4f;
```
```
    short z = 1;
```

    Console. WriteLine((float) x + y * z - (x + = (short) y) );

```
    Console. ReadLine();
```
```
}
```

Answer: d
Explanation: None.
Output :

0.4000001

10. A float occupies 4 bytes. If the hexadecimal equivalent of these 4 bytes are A, B, C and D, then when this float is stored in memory in which of the following order do these bytes gets stored?
a) ABCD
b) DCBA
c) 0 * ABCD
d) Depends on big endian or little endian architecture

Answer: d
Explanation: “Little Endian” means that the lower-order byte of the number is stored in memory at the lowest address, and the high order byte at the highest address. For example, a 4 byte Integer
ABCD will be arranged in memory as follows:
Base Address + 0 Byte 0.
Base Address + 1 Byte 1.
Base Address + 2 Byte 2.
Base Address + 3 Byte 3.
Intel processors (those used in PC’s) use “Little Endian” byte order.
“Big Endian” means that the high-order byte of the number is stored in memory at the lowest address, and the low-order byte at the highest address. The same 4 byte integer would be stored as:
Base Address + 0 Byte 3.
Base Address + 1 Byte 2.
Base Address + 2 Byte 1.
Base Address + 3 Byte 0.

```
 public static void Main(string[] args)
```
```
 {
```
```
     double ZERO = 0;
```

     Console.WriteLine("RESULT OF DIVISION BY ZERO IS :{0}",  (0 / ZERO));

```
     Console.ReadLine();
```
```
 }
```

13. Which of the following format specifiers is used to print hexadecimal values and return value of output as Octal equivalent in C#?
a) %hx for small case letters and %HX for capital letters
b) %x for small case letters and %X for capital letters
c) No ease of doing it. C# don’t provides specifier like %x or %O to be used with ReadLine() OR WriteLine(). We have to write our own function
d) %Ox for small case letters and %OX for capital letters

C# Questions & Answers – Floating and Decimal Data Types