SOLUTION

There are four types of scopes provided by C++: 
1.  Local scope: A local variable scope is restricted to the function that declares the variable.
2.  Function scope: The variables declared in the outermost block of a function have function scope.
3.  File scope: A name declared outside all blocks and functions has file scope.
4.  Class scope: A name of a member has class scope and is local to its class.

SOLUTION

  The types of functions provided by C++ are: 
1.  User defined functions.
2.  Built in functions.

SOLUTION

A function declaration establishes the name of the function and the number and type of its parameters. A function declaration consists of a return type, a name and a parameter list.
A function definition contains a function declaration and body of the function.

SOLUTION

Global variables are the variables that are declared outside all functions. These variables can be accessed from outside the function, whereas local variables are the variables which are declared within the body of the function or a block and can be accessed by these functions only.

SOLUTION

Scope are the rules that decide in which part of the program a particular piece of code or data item would be known and can be accessed therein.

SOLUTION

Whenever a C++ program is executed, execution of the program starts at main (). The main is the driver function of a program. If it is not present in the program, no execution can take place.

SOLUTION

When a function prototype appears before the function definition, it is known as global prototyping.

SOLUTION

Prototyping is required as compiler carefully compares each use of function with its prototype to check whether the function is invoked properly.

SOLUTION

i) int val(int n1, int n);
ii) float sum(float a);

SOLUTION

datatype function name (parameter list)
{ 

body of the function
}

SOLUTION

User defined functions are created by the programmer as per the requirement of the program.

SOLUTION

Function Main( ) primarily consist of:
a) The data items required and
b) The function calls.

SOLUTION

A function in C++ language is a block of code that performs a specific task. It has a name and it is reusable i.e. it can be executed from as many different parts in a C++ program.

SOLUTION

In C++ a function may return the members of a structure , Only one value can be returned by a function, either the whole structure or just one member. Here we can pass a structure to a function through call by value as well as call by referenc mechanism.

SOLUTION

The prototype void fun(ABC &, char[]); indicates that arguments are passed by reference. The first argument represents that object is passed as a reference and the second argument is an array of character which by default works on reference. 

SOLUTION

We can tell how variables are being sent through the function prototype or definition. The function call is the same for call by value and by reference.

SOLUTION

The #define is a preprocessor directive that allows us to define symbolic names and constants.

SOLUTION

In C++ a structure can be passed to a function through "call by value" as well as "call by reference" mechanism. So if we pass a structure through call by reference then the original value can also get changed.

SOLUTION

A structure is passed as a value to avoid changes in the actual object during any changes done in the formal argument in a function body.

SOLUTION

The given code has been executed in such a way - #include< iostream.h > 
struct Pt
{
   int x,y;
};
void disp(Pt p)
{
  cout << " "<< p.x <<" " << p.y;  
}
void main()
{
 Pt A,B,C={25,50};//values of x and y for A,B and C structure variable
 A=C;                //values of x and y of C assigned to the x and y of A
 C.x+=5;         //x of C =30 
 B=C;               //values of x and y of C assigned to B
 B.y+=10;      // y of B = 60
 A.x=22;        //x of A = 22
 disp(A);        // calling disp() function and A has passed into it 
 disp(B);       // calling disp() function and B has passed into it   disp(C);       // calling disp() function and C has passed into it 
}   The output of the code is 22 50 30 60 30 50

SOLUTION

To access the third element of the list we can use the statement- NE[2].code and NE[2].name

SOLUTION

EMP[45].salary.tax statement is an array of nested structure.

SOLUTION

A # sign in the beginning of a statement means, it is a preprocessor directive, i.e., the instructions given here will be processed before the program is compiled.

SOLUTION

A structure inside another structure is called as nested structure.

SOLUTION

For ABC, a reference variable is being used while for char, it is being passed by value.

SOLUTION

typedef salary amount cannot be used because salary is not a fundamental data type.

SOLUTION

Macros always replace the value of a variable before compilation, while const acts like a simple constant during the compilation.

SOLUTION

A self-referential structure has an element that, refers to the structure itself. This is actually a pointer that points to a structure of the same type.

SOLUTION

Tag is the name of the structure, which is being declared after the keyword struct.

SOLUTION

The pointer gets incremented by the size of the structure.

SOLUTION

A structure can contain only the data while both data and functions are included in a class.

SOLUTION

A structure is a heterogeneous data structure since, it stores different types of data.

SOLUTION

var is a pointer , so to access it b- > var is used.

SOLUTION

The functions that calculate or compute some value and return the computed value are computational functions.

SOLUTION

In call by reference, instead of passing a value to the function being called, a reference to the original variable is passed.

SOLUTION

extern is used when we have defined the variable elsewhere, and use that variable in our program.

SOLUTION

In call by reference, the changes are reflected back to the original values.

SOLUTION

Functions that perform an action and have no explicit return value are procedural functions.

SOLUTION

When a function is called by reference, then the formal parameters become references to the actual parameters in the calling function.

SOLUTION

In above function as type and number of parameters are not specified so it is considered as open parameter list.

SOLUTION

Arrays and pointers are derived data types. an array can hold multiple elements of same data type.

SOLUTION

A function is invoked by providing its function name, followed by the parameters.

SOLUTION

‘Void’ data type is used for empty set of values.

SOLUTION

No, constants or expressions cannot be passed by reference.

SOLUTION

 User defined functions are created by the programmer as per the requirement of the program.

SOLUTION

 A prototype has no body and no code.In other words, a (prototype) declaration introduces a function name to the program. On the other hand, a definition tells the Program, what is the function doing and how is it doing so.

SOLUTION

 return-type function name (parameter list)

}

SOLUTION

 A function is a subprogram that acts on data and often returns a value.

SOLUTION

 The two main parameters that appears while calling a function are:
1. Actual parameter 
2. Formal parameter

SOLUTION

The keyword that is used to make an argument constant is ‘Constant’.

SOLUTION

Yes, an argument can have default values.

SOLUTION

Syntax: void function-name (parameter list);

SOLUTION

The place of function or variable declaration determines its scope.

SOLUTION

Structure is a user define data structure that is a collection of various types of data elements. The struct keyword is used to declare the structure.

SOLUTION

The C++ preprocessor is a program that is executed before the source code is compiled. It always begins with a hash "#" and ends without a semicolon ";"

SOLUTION

name is the member of the structure "employee". so to access any member of a structure we need to use structure variable with the dot operator. So the correct statement will be - cout << emp.name;

SOLUTION

stu_name and stu_sex are the members of the structure "Student". so to access it we need to use structure variable i.e. "st". In a structure we can declare the members only but we cannot initialize it.

SOLUTION

If declared within main() it becomes local to the function main().  

SOLUTION

The program will execute as follows - #include < iostream.h >
struct Sumval
{
int a,b;
}
void Struct_Fun(Sumval Sv)
{
Sv.b = 2.5;
cout<< Sv.a << Sv.b;
}
void main ( )
{
Sumval Sv;
Sv.a=100;
Sv.b=1.5;
Struct_Fun(Sv); // calling Struct_Fun function
cout<< Sv.a << Sv.b;
}   The output will be - 10021001

SOLUTION

C++ allows us to define explicitly new data type names by using the keyword typedef. The typedef does not actually create a new data class, rather it defines a new name for an existing type. It can be used with any data type.

SOLUTION

C++ allows us to define explicitly new data type names by using the keyword typedef. Using typedef does not actually create a new data class, rather it defines a new name for an existing type. The syntax to create an alias for the int data type is - typedef int id; Here id is an alias for int data type.

SOLUTION

The statement "typedef salary amount;" will generate an error as salary is not a data type. Here we can use only the fundamental or basic data types of C++.

SOLUTION

 An array always has elements of the same data type  

SOLUTION

The syntax to access the members of the structure is - structure-variable.member-name  

SOLUTION

Using typedef keyword we can make the code easier to read and understand.

SOLUTION

vi, values cannot be assigned this way to a declared structure  variable.

SOLUTION

In C++ we can declare members of a structure only. we cannot initialize it.

SOLUTION

a1.x  or  a2->x, a2 is a pointer variable hence uses -> to access the member

SOLUTION

vb is the variable of B structure and mn is the nested variable of A structure. The statement to access r which is the member of B structure is - vb.r The statement to access x which is the member of A structure is - vb.mn.x

SOLUTION

An alias for structure tag name can be defined using typedef keyword. It defines another name for the structure tag name.

SOLUTION

As p1 is the structure variable of person and p is the structure variable of name. The correct statement to display the name is - cout<

SOLUTION

The preprocessing phase of a C++ program occurs before a program is compiled. The C++ preprocessor is a program that is executed before the source code is compiled.

SOLUTION

macro replaces value before compilation while const will declare a variable during compilation

SOLUTION

v1=v2 will work only, when both are of the same data type.

SOLUTION

If declared within main(), it becomes local to the function main().

SOLUTION

typedef is used to give an alias name to the existing data types.

SOLUTION

Since, the structure variable belong to the structure data type. Therefore, it has to be passed in the form of a structure.

SOLUTION

gets( stu_name); because methods are accessed with dot operator.
int age=17; because, we cannot initialize variables within a structure.

SOLUTION

it should be cout << emp.name.

SOLUTION

The dot operator has a structure variable on the left and structure member on the right side.

SOLUTION

The syntax for structure declaration is :     struct structure_name {
   member_type1 member_name1;
   member_type2 member_name2;
   member_type3 member_name3;
    .
    .
    } object_names;

SOLUTION

Any array starts from index 0. So, the third employee’s details can be obtained by NE[2].name and NE[2].code.

SOLUTION

Variables are never initialized during a structure declaration.

SOLUTION

r is the member of structure B and x is a member of structure A, which has been nested inside B.

SOLUTION

A value can never be assigned inside a structure.

SOLUTION

P1 is an instance of structure person which, includes name which in itself is another structure. So, complete access can be done by the statement 
       P1.p.title, P1.p.fname, P1.p.lname

SOLUTION

The sequence of calling functions is : disp (A); disp (B); disp (C);, and the output is displayed according to this sequence.

SOLUTION

A semicolon should always terminate a structure.

SOLUTION

A function may return the members of a structure, Only one value can be returned by a function, either the whole structure or just one member.

SOLUTION

When tag is missing, only one variable can be declared. Here, three have been declared and this is wrong.

SOLUTION

While declaring a structure without a tag, only one variable can be declared.

SOLUTION

typedef defines new name in addition to the existing type name. It does not replace the standard data type name.

SOLUTION

Nested structures contain structures within structures. A structure element may be either complex or simple. The simple elements are any of the fundamental data types of C++ i.e., int, char, float, double. However, a structure may consist of an element that itself is complex i.e., it is made up of fundamental types, e.g., arrays, structures etc.

SOLUTION

The total number of elements of a two-dimensional array is get calculated by using the formula - Total_no_of_elements = no_of_rows x no_of_columns. So here the total number of elements is 30.

SOLUTION

An array marks of type int with 35 elements will have marks[0] at the first allocated memory location with last subscripts as 34.

SOLUTION

A string is defined as a character array that is terminated by a null character. Therefore, the char array are declared one character longer to hold null at the end.

SOLUTION

Two-dimensional array is the simplest form of multi-dimensional array. It has two dimensions, one represents the row and another represents the column.

SOLUTION

The syntax of an array declaration is - type array-name[size], Here "type" declares the base type of the array, which is the type of each elements of the array.

SOLUTION

An array is a collection of variables of the same type that are referenced by a common name. It is of two type i.e. single-dimensional array and multi-dimensional array.

SOLUTION

Subscripts is used either to retrieve the value of one or more array elements or to allow array elements to receive new values.