Chapter 22, Problem 22.4E

Program Plan Intro

To Determine:

Definition for each of the following:

1. Structure inventory, containing character array part name [30], integer part number, floating-point price, integer stock and integer reorder.
2. A structure called address that contains character arrays street address [25], city [20], state [3] and zipcode [6].
3. Structure student, containing arrays first name [15] and last name [15] and variable home address of type struct address from part (b).
4. Structure test, containing 16 bit fields widths of 1 bit. The names of the bit fields are the letters a to p.

Explanation of Solution

Explanation:

Defining Structures:

Structure is nothing but the collection of different variables of different types. First we need to define a structure in the program, for that we need to use the keyword struct and then the name of the structure. The name under which we want to combine different-different variables or different-different elements and between the curly braces whatever we write or whatever variables we declare will be considered as the members of the structure. All the members will be available under this one name of the structure. Declaring or defining the member of a structure is similar to the way we declare the variables.

1. Here we are storing the information of Inventory, so the structure name will be Inventory and all the members are available under this one name Inventory.

structInventory
{
	charpartName[30];
	intpartNumber;
	floatprice;
	intstock;
	intrecorder;
};

2. Here we are storing the information of Address, so the structure name will be Address and all the members are available under this one name Address.

structAddress
{
	charstreetAddress[25];
	charcity[20];
	charstate[3];
	charzipCode[6];	
};

3. Here we are storing the information of Student, so the structure name will be Student and all the members are available under this one name Student.

structStudent
{
	charfirstName[15];
	charlastName[15];
	struct
	{
		charstreetAddress[25];
		charcity[20];
		charstate[3];
		charzipCode[6];
	}
	homeAddress;
};

4. Here we are storing the information of Test, so the structure name will be Test and all the members are available under this one name Test.

structTest
{
	boola:1, b:1, c:1, d:1, e:1, f:1, g:1, h:1,
	i:1, j:1, k:1, l:1, m:1, n:1, o:1, p:1;
};