CS100 : Computer Programming

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1 CS100 : Computer Programming Chapter 6 - Structures, Unions and Enumerations Narasimhan T. 6.1 Structures Suppose you want to store the details of a book. You may have to store details like title, price and edition. What is the big deal? You just define three variables viz. title, price and edition. This seems perfectly fine until you think more on it. The three pieces of information : title, price and edition are related in the sense that they all pertain to a single book. But the corresponding variables are unrelated : they are three separate entities with no relation among them. You need a technique to relate the variables or wrap these o a single unit. You cannot use an array in the above situation as the three variables are of different types. You need a more powerful user defined data type in which the individual elements can differ in type. To solve this problem, C provides a special data type the ure. A ure is a collection of one or more variables (elements), possibly of different types, grouped together under a single name for convenient handling. The individual ure elements are referred to as members or fields. Structures help to organize complicated data, particularly in large programs, because they permit a group of related variables to be treated as a single unit instead of as separate entities Defining a ure A ure must be defined in terms of its individual members. In general terms, a ure may be defined as: ; tag member-1; member-2;.. member-m; In this declaration, is a required keyword; tag is a name that identifies the ure; and member-1, member-2,, member-m are the member declarations. Notice that the declaration is terminated by a semicolon. This is because a ure declaration is a statement. (There is no formal distinction between a ure definition and a ure declaration; the terms are used erchangeably.) 1

2 Example 6.1. For the book scenario discussed earlier, we could define a ure like this ; float book title[15]; price; edition; Once the new ure data type has been defined, one or more variables can be declared to be of that type. This is done as follows: tag variable-1, variable-2,..., variable-n; where is a required keyword, tag is the name that appeared in the ure declaration and variable-1, variable-2,..., variable-n are ure variables of type tag. Example 6.2. This example declares three variables b1, b2, b3 of the type book. book b1, b2, b3 ; To avoid the use of the keyword in variable declarations, we can use as illustrated below. Example 6.3. This is another way to declare ure variables. title[15]; float price; edition; book; book b1,b2; It is also possible to combine the ure definition and ure variable declaration in one statement. Example 6.4. This is yet another way to declare ure variables. book title[15]; float price; edition; b1,b2,b3; 6.2 Accessing ure elements Individual members of a ure are accessed through the use of the dot ( ) operator. For example, to pr the edition of book b1, we write for more log on to 2

3 prf("%d",b1.edition); Similarly to input the title of book b3, you write: gets(b3.title); or scanf("%s",b3.title); Example 6.5. This program inputs and displays the details of a book. #include<stdio.h> book title[15]; author[15]; edition; float price; ; main() book b; prf("enter the title of the book\n"); gets(b.title); prf("enter the author of the book\n"); gets(b.author); prf("enter the price of the book\n"); scanf("%f",&b.price); prf("enter the edition of the book\n"); scanf("%d",&b.edition); prf("book DETAILS\n"); prf("title : %s\n",b.title); prf("author : %s\n",b.author); prf("edition : %d\n",b.edition); prf("price : %f\n",b.price); 6.3 Initializing ure members The members of a ure variable can be assigned initial values in much the same manner as the elements of an array. The initial values must appear in the order in which they will be assigned to their corresponding ure members, enclosed in braces and separated by commas. Example 6.6. This example illustrates how a ure variable is initialized. float book title[15]; price; 3

4 ; edition; book b="c Programming", , 2; This statement assigns C Programming to title, to price and 2 to edition. It is also possible to initialize the ure variable in the ure definition. Example 6.7. This is another way to initialize ure variables. book title[15]; float price; edition; b="c Programming", , 2; There are few rules to keep in mind while initializing ure variables in the program. 1. You cannot initialize individual members inside the ure template. 2. The order of values enclosed in braces must match the order of members in the ure definition. 3. It is possible to initialize only the first few members and leave the remaining uninitialized. The uninitialized members should be only at the end of the list. 4. The uninitialized members will be assigned default values as follows: Zero for eger and floating po numbers \0 for acters and strings 6.4 Operations on ure members and on ure variables The individual members in a ure can be manipulated using any operation. Following are some examples: if(b.price>100) b.price-=10; b.price++ The precedence of the member operator (. ) is higher than all arithmetic and relational operators. So there is no need to put parenthesis like (b.price)++ You can assign one ure variable to another variable of same ure. You need not assign the value of each member separately. Example 6.8. This program illustrates ure assignments for more log on to 4

5 sample a, b; x, y; x.a = 10; x.b = 20; y = x; prf("%d %d", y.a, y.b); This will pr Comparison of two ure variables is not permitted. Following statements will result in errors. if(b1==b2) if(b1!=b2) However you can compare two ure members like if(b1.price>b2.price) b1.price-=50; 6.5 Array of ures Just as you have array of egers or of acters, you can also define an array of ure variables. Such an array is called array of ures an array in which each element is a ure variable. This is illustrated in the following example. Example 6.9. This program inputs and displays the details of several books. #include<stdio.h> book title[15]; author[15]; edition; float price; ; main() book b[10]; i,n; prf("enter how many books you want to process\n"); scanf("%d",&n); for(i=0;i<n;i++) prf("enter the title of book %d \n",i+1); scanf(" %s",b[i].title); 5

6 prf("enter the author of book %d \n",i+1); scanf(" %s",b[i].author); prf("enter the edition of book %d \n",i+1); scanf("%d",&b[i].edition); prf("enter the price of book %d \n",i+1); scanf("%f",&b[i].price); for(i=0;i<n;i++) prf("book %d DETAILS\n",i+1); prf("title : %s\n",b[i].title); prf("author : %s\n",b[i].author); prf("edition : %d\n",b[i].edition); prf("price : %f\n",b[i].price); 6.6 Nested ures Suppose we have a ure named employee with the following template: ; employee empname[15]; empid; housename[15]; streetname[15]; cityname[15]; pincode; state[15]; salary; Here the members housename to state collectively represent the address of the employee. Thus they could conveniently be combined to form a ure as follows: ; employee empname[15]; empid; address housename[15]; streetname[15]; cityname[15]; pincode; state[15]; addr; salary; for more log on to 6

7 ; The above template can also be written as: ; address housename[15]; streetname[15]; cityname[15]; pincode; state[15]; employee empname[15]; empid; address addr; salary; To access the housename we write emp.addr.housename where emp is the variable of type employee. It is also possible to nest more than one ure inside another ure. 6.7 Unions Union is another way to group together data whose types may differ from one another. However, all the members within a union share the same storage area within the computer s memory, whereas each member within a ure is assigned its own unique storage area. Thus, unions are used to conserve memory. They are useful for applications involving multiple members, where values need not be assigned to all of the members at any one time. However, since all the union members share the same location, only one member can be accessed at a time. Declaring a union is similar to declaring a ure. Its general form is: union ; tag member-1; member-2;.. member-m; where union is a required keyword and the other terms have the same meaning as in a ure definition. Once a union is defined, union variables can then be declared as: union tag variable-1, variable-2,..., variable-n; where union is a required keyword, tag is the name that appeared in the union definition and variable-1, variable-2,..., variable-n are variables of type tag. Example Consider the definition 7

8 union cp; result grade; marks; Here you have a union variable, cp, of type result. It can represent either a acter(grade) or an eger quantity (marks) at a time. To access a union member, you use the same syntax as that you did for ure members. For example, with the union definition as above, you could write cp.grade= S ; While accessing an union member, you should make sure that you are accessing the member whose value is currently stored. For example, the code cp.grade= S ; cp.marks=92; prf("you have obtained %c grade for end semester examination in Computer Programming",cp.grade); could yield unexpected results. This is because, with the union definition as earlier, you can store only one value, either grade or marks in the allocated space. Thus when you store grade and later try to store marks, it will overwrite grade. 6.8 Enumerations Enumeration is a user defined data type. It can be used to define new data types. An enumeration may be defined as enum identifier value-1, value-2,..., value-m; where enum is a required keyword; identifier is a name that identifies the enumeration, and value-1, value-2,..., value-m represent the possible values that any variable of the defined type can take. These possible values are known as enumerators or enumeration constants. Once the enumeration has been defined, corresponding enumeration variables can declared as enum identifier var-1, var-2,..., var-n; where identifier is the name that appeared in enumeration definition and var-1, var-2,..., var-n are variables of the defined type. Example A C program contains the following declarations. enum enum colours black, blue, cyan, green, magenta); colours foreground, background; The first line defines an enumeration named colours (i.e., we define a new data type colours). Any variable of colours type can take one of the possible values black, blue, cyan, green, magenta. for more log on to 8

9 The second line declares two variables foreground and background to be enumeration variables of type colours. Thus, each variable can be assigned any one of the values black through magenta. The two declarations can be combined if desired, resulting in enum colours black, blue, cyan, green, magentaforeground, background; Enumeration constants are automatically assigned equivalent eger values, beginning with 0 for the first enumerator, and for each successive enumerator, the assigned value being increased by 1. Example Consider the enumeration defined in Example The enumeration constants will represent the following eger values. black 0 blue 1 cyan 2 green 3 magenta 4 It is possibe for the user to specify values of his choice for enumerators. See the example below: Example Consider the enumeration defined as below: enum colours black=-1, blue, cyan, green=0, magenta); This results in the assignment of values as follows: black -1 blue 0 cyan 1 green 0 magenta 1 Note that blue and green both are assigned 0. Similarly both cyan and magenta have the value 1. Enumeration variables can be processed in the same manner as other eger variables. Thus, they can be assigned new values, compared, etc. With the definition as in Example 6.11, following statements are valid. foreground = magenta; if(background == blue) foreground = cyan; if(background == foreground) prf("background and forefround are of same colour"); 9

10 switch(background) case blue: foreground=black; break; default: prf("bye!"); The key po to understand about an enumeration is that each of the enumerators stands for an eger value. As such, the enumeration constants cannot be pred, only their eger values can be pred. Consider the code fragment: enum colours black, blue, cyan, green, magentaforeground, background; foreground = cyan; prf("%s",foreground); This will result in an error. This is because, the enumerators are not strings (although, they appear to be). They are just names for egers. Internally they are treated as egers only. Thus to pr the equivalent eger value, you could modify the last statement as prf("%d",foreground); and this will pr You can define new data type by using the keyword. You are not actually creating a new data type, but rather defining a new name for an existing type. In general terms, a new data type is defined as type new-type; where type refers to an existing data type (either a standard data type like, etc, or previous user-defined data type), and new-type refers to the new user defined data type. Example Here is a simple declaration involving the use of. float marks; In this declaration marks is an user-defined data type, which is equivalent to type float. This statement tells the compiler to recognize marks as another name for float. Once a new name has been defined for an existing type, then new variables, arrays, etc. can be declared in terms of this new name. For example, to store the marks you scored for Computer Programming, you could write marks cp; This is equivalent to writing float cp; for more log on to 10

11 If you want to store the marks of 74 students, you could just write marks cp[74]; Using can make your code easier to read and to understand. This feature is particularly convenient when defining ures, since it eliminates the need to repeatedly use the keyword whenever a ure is referenced. Hence, the ure can be referenced more concisely. See Example Programming examples Program 6.1. Input the details (name, roll number and marks in 6 subjects) of n students and pr the details along with the total marks scored. #include<stdio.h> main() name[15]; roll, marks[6], total; student; student s[10]; // You assume n is less than or equal to 10 n,i,j; prf("how many students?"); scanf("%d",&n); for(i=0;i<n;i++) prf("enter student name\n"); scanf("%s",s[i].name); prf("enter roll number\n"); scanf("%d",&s[i].roll); prf("enter marks in six subjects\n"); for(j=0;j<6;j++) scanf("%d",&s[i].marks[j]); s[i].total=0; for(j=0;j<6;j++) // You could find the total in the previous loop too. s[i].total+=s[i].marks[j]; for(i=0;i<n;i++) prf("student %d DETAILS\n",i+1); prf("name:%s\n",s[i].name); prf("roll:%d\n",s[i].roll); for(j=0;j<6;j++) prf("marks %d: %d\n",j+1,s[i].marks[j]); prf("total:%d\n",s[i].total); 11

12 Program 6.2. Input the details (name, employee ID and date of joining) of two employees. Represent the date of joining as a nested ure. Pr the details of the employee with more experience in the company. Assume the two employees have joined in different years. #include<stdio.h> doj; doj d; employee; date; month[15]; year; name[15]; empid; main() employee e[2]; i; for(i=0;i<2;i++) prf("enter employee name\n"); scanf("%s",e[i].name); prf("enter employee ID\n"); scanf("%d",&e[i].empid); prf("enter day of joining\n"); scanf("%d",&e[i].d.date); prf("enter month of joining\n"); scanf("%s",e[i].d.month); prf("enter year of joining\n"); scanf("%d",&e[i].d.year); prf("details OF THE SENIOR EMPLOYEE\n"); if(e[0].d.year<e[1].d.year) prf("name:%s\n",e[0].name); prf("employee ID:%d\n",e[0].empID); prf("date of joining:%d %s %d\n",e[0].d.date,e[0].d.month,e[0].d.year); else prf("name:%s\n",e[1].name); prf("employee ID:%d\n",e[1].empID); prf("date of joining:%s %d %d\n",e[1].d.month,e[1].d.date,e[1].d.year); for more log on to 12

13 Program 6.3. In a company there are two type of employees Type0 and Type1. Employees belonging to Type0 are paid on hourly basis at the rate of 750 per hour. Type1 employees are paid monthly, and their gross salary is calculated as the sum of basic pay, DA (125% of basic pay) and HRA ( 550). Input the number of hours for Type0 employees and basic pay of Type1 employees and pr the gross salary. #include<stdio.h> union hours; float basicpay; salary; float salary s; employee; name[15],dept[15]; type; gross; main() employee emp; prf("enter the name of employee"); scanf("%s",emp.name); prf("enter the department of employee"); scanf("%s",emp.dept); prf("enter the type of employee:0 or 1\n"); scanf("%d",&emp.type); if(emp.type==0) prf("enter the number of hours\n"); scanf("%d",&emp.s.hours); emp.gross=emp.s.hours*750; else prf("enter the basic pay\n"); scanf("%f",&emp.s.basicpay); emp.gross=emp.s.basicpay+emp.s.basicpay* ; prf("employee Details\n"); prf("name:%s\n",emp.name); prf("department:%s\n",emp.dept); prf("salary:%f\n",emp.gross); 13