Q3. What is
difference between dynamic and reference declaration of variable?
Answer: Dynamic initialization is extensively used
in object-oriented programming. We can create exactly the type of object needed
using information that is known only at the run time.
C++
introduces a new kind of variable known as the reference variable. A reference
variable provides an alias (alternative name) for a previously defined
variable. For example, if we make the variable sum a reference to the variable
total, then sum and total can be used interchangeably to represent the
variable.
Q4.What is type
casting? explain the use of type casting with suitable example.
When there are two operands of
different types in an expression, the lower-type variable is converted to the
type of the higher-type variable. For example, if one operand is of type int
and other is of float, both are converted into float type. Thus,
if data types are mixed in an expression, C++ performs the conversions
automatically. This process is called implicit or automatic conversion.
An example:
void
main()
{
int a=10;
float b=40.5, result;
result=a+b;
cout<<"The sum is
"<<result; }
Output:
The sum is 50.5.
In this case, variable a
is of type int while b is of type float. While solving the expression a+b,
the temporary variable of type float is created to store value of a and then
two float values are added. The result is of type float and stored in variable
result. Here, data type int is promoted to float. Thus, it is also known as standard
data type promotion.
The other implicit conversions are
Int and float -> float
Int and long -> long
Int and double -> double
Long and float -> float
Float and double -> double
But,
sometime programmer needs to convert a value from one type to another
explicitly in a situation where the compiler will not do it automatically. This
type of conversion is called explicit conversion or type casting.
An example:
void main()
{
int a=4000;
long
result;
result=a*100;
cout<<"The product is
"<<result; }
Output:
The product is 6784 // Error!!!!!
The
corrected version is:
void main()
{
int a=4000;
long
result;
result=long(a)*100; //type casting
cout<<"The product is
"<<result; }
Output:
The result is 400000 // correct
Here, while solving the expression
long(a)*100 in above program, integer type variable a is
converted into long type. For this, a temporary variable of type long is
created to store value of a and this long value is multiplied by 100 and
result is also of type long. The long result is stored in variable result.
Without casting, the result is
not correct. This is because multiplication of two integers (i.e. a and 100)
results integer value (i.e. 4000000). But this integer value is beyond the
range of integer. Thus, garbage value is stored in result variable.
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