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CHAPTER-13 STRUCTURES SHORT ANSWER QUESTIONS 1. Discuss the similarity and differences of a structure as compared to an array and a class. Ans. Similarities array and Structure Both structures and arrays are used to hold more than 1 elements under single name. Difference array structure Arrays bring together a group of items of the same data type Structures bring together a group of related data items of any data type. Similarities class and structure Both Class and Structures can have methods, variables and objects. Both can have constructor. Both are user defined types. Difference class structure Declared with the keyword class Declared with the keyword struct By default, all members are private in a class By default, all members are public in a structure 2. Define a structure to store information about a fish. The structure should include the kind, the weight in grams, and the length in inches. Declares variables of this structure type and discuss various methods of initializing them. Ans. struct Fish_Info { char kind[50]; int gram; int length; }; Fish_Info fish; first method of initializing The structure elements of a structure can be initialized separately, using separate assignment statement. fish.kind=abc; fish.gram=300; fish.length=5; second method of initializing The structure element of a structure can be initialized jointly, using the notation similar to array initialization. Fish_Info fish = { abc, 300, 5); 3. Write definition for a structure EMPREC that stores information about an employee such as empno, name, address, salary, and joining_date. The address member of EMPREC stores the information houseno, area, and city. The joining_date member of EMPREC stores information day, month, and year. Ans. struct date { int day; int month; int year; }; struct addr 1 { int houseno; char area[31]; char city[31]; }; struct EMPREC { int empno; char name[41]; addr address; float salary; date joining_date; }; 4. What is structure? Declare a structure in C++ with name, roll number and total marks as components. Ans. A structure is a collection of variables having different data types. The declaration as: struct student { char name[20]; int roll_no,marks; }; 5. Rewrite the following program after removing the syntactical error(s), if any. Underline each correction. #include int main( ) { struct STUDENT { char stu_name[20]; char stu_sex; int stu_age=17; }student; gets(stu_name); gets(stu_sex); return 0; } Ans. #include #include int main( ) { struct STUDENT { char stu_name[20]; char stu_sex; int stu_age; //Initialization of variables inside a structure is not allowed. }student; gets(student.stu_name); cin>>student.stu_sex; //A single character cannot be read using gets return 0; } 6. What are Nested Structures? Give an example. Ans. Nested structures are structures as member of another structure. For example,the date of birth is astructure within the structure of a student as shown below. These types of structures are known as nested structures. 2 Example: - #include< iostream.h> struct course { int couno; int coufees; }; struct student { int studno; course sc; course sc1; }; void main( ) { student s1; s1.studno=100; s1.sc.couno=123; s1.sc.coufees=5000; s1.sc1.couno=200; s1.sc1.coufees=5000; int x = s1.sc.coufees + s1.sc1.coufees; coutmemnumber.per1; coutper1.membername; per1.memtype = HIG; per2=per1; cin>>Member Number; cinper1.memnumber; coutper1.memname; strcpy(per1.memtype, HIG); per2=per1; cout>>Member Number; : ss1=ss3; } Suggest way(s) to rectify the errors. Ans. #include struct s1 { int a; float b; char c; }st1,st2,st3; int main() { struct s2{ int x; float y; char z; }ss1,ss2,ss3; // Read & Initialize structures ss2=ss1; : ss3.z=st1.c; : } void func1() { ss2.x=st1.a; ss3.y=st2.b; ss1.z=st3.c; : ss1=ss3; } The error points mentioned in this answer is depend on the compiler, as we are using stone age compiler this may give compilation the error, but the ss3 will show error in all complier because ss3 is not declare inside func1(). 9. Write a code fragment to declare and read in values (from user) for an array ARR of size 10 whose elements consists of name, category, Marks in 5 subjects and registration number. The category can be one of Gen, SC, ST, and OBC. The registration number is a combination of areacode, region code, school code, current year (yy) and a serial number. Ans. #include #include #include struct rege_num { int areacode; int region_code; int school_code; int current_year; int serial_number; }; struct info { char name[50]; char category[10]; int marks[5]; 5 rege_num regno; }; info arr[10]; void main() { int i,j; clrscr(); for(i=0;i int phone_no; }; struct book_list { int bookno; char bookname[50]; char subject[20]; float price; char edition[10]; publication public[20]; }; struct book_info { int authorno; char author_name[50]; book_list book[5]; }; 11. Write a code fragment to read in values for the first record of array mentioned in question 10. Ans. #include #include #include struct publication { char proprietor_name[50]; char company_name[50]; char address[50]; int phone_no; }; struct book_list { int bookno; char bookname[50]; char subject[20]; float price; char edition[10]; publication pub; }; struct book_info { int authorno; char author_name[50]; book_list bookl[5]; }; book_info book[2]; void main() { int i,j; clrscr(); for(i=0;i cin>>book[i].bookl[j].bookno; cout int m = d1.mm+(d/30); d1.mm=m%12; d1.yy += (m/12); if(d1.dd= =0) { d1.dd=12; --d1.mm; } if(d1.mm< =0) { d1.mm=30; --d1.yy; } return(d1); } 13. Predict the output of the following code (Assuming that required header files have been included): struct Time { int hour, minutes, seconds; }; void updatetime(Time new_time) { ++new_time.seconds; if(new_time.seconds60) { new_time.seconds=0; ++new_time.minutes; } if(new_time.minutes60) { new_time.minutes=0; ++new_time.hour; } if(new_time.hour24) new_time.hour=0; } int main() { Time cur_time; coutcur_time.hour>>cur_time.minutes>>cur_time.seconds; updatetime(cur_time); cout #include struct Package { int Length,Breadth,Height; }; void Occupies(Package M) { cout } Ans. 16. What will the output of the following code, if the input entered is: 1.9, 2.1, -3.4, 2.1, 2.2, -4.5, 3.2, 2.6, 2.9, 23.1, -4.4, 3.1, 3.2, -3.5, 2.2, 1.6 #define LOOPS 4 typedef struct imaginary_number { double real_part; double imag_part; }IMAG_NUM; typedef struct imag_rect { IMAG_NUM corner1; IMAGE_NUM corner2; }IMAG_RECT; IMAG_RECT Rect[3]; Rect[i].corner1.imag_part=3.1; Rect[i].corner1.real_part=1.2; Rect[i].corner2.imag_part=-2.3; Rect[i].corner2.real_part=1.4; int main() { int i,j; for(i=1;i>Rect[i].corner1.imag_part; cin>>Rect[i].corner1.real_part; cin>>Rect[i].corner2.imag_part; cin>>Rect[i].corner2.real_part; } for(i=0;i #define LOOPS 4 typedef struct imaginary_number { double real_part; double imag_part; }IMAG_NUM; typedef struct imag_rect { IMAG_NUM corner1; IMAG_NUM corner2; }IMAG_RECT; IMAG_RECT Rect[4]; int main() { int i,j; clrscr(); for(i=0;i>Rect[i].corner1.imag_part; cin>>Rect[i].corner1.real_part; cin>>Rect[i].corner2.imag_part; cin>>Rect[i].corner2.real_part; } for(i=0;i { cm n1,n2,res; cout int runs; int wicket; }; void main() { batting bat_team[12]; bowling bo_team[12]; cout if(bat_team[i].out) cout int empno; char name[21]; float basic; info others; } salemp, puremp, imemp; void main() { float gs, ns; clrscr(); cout salemp.basic>> salemp.others.hra_rate >> salemp.others.da_rate >> salemp.others.cca >> salemp.others.pf_rate >> salemp.others.it_rate; gs = (salemp.basic * salemp.others.days_worked) /26 + (salemp.basic * salemp.others.hra_rate)/100 + (salemp.basic * salemp.others.da_rate)/100 + salemp.others.cca; ns = gs-((salemp.basic * salemp.others.it_rate)/100 +(salemp. basic * salemp.others.pf_rate)/100); cout { int area_code; int exchange; int number; }; struct member { int member_id; char name[30]; phone ph_no; }; void main() { member rec[20]; char ch; int i; cout> ch >> rec[i].ph_no.exchange >> ch >> rec[i].ph_no.number ; } clrscr(); cout cout >room.b_in_feet >> ch >>room.b_in_inch >> ch; cout >room.h_in_feet >> ch >>room.h_in_inch >> ch; float len, br, height; len =room.l_in_feet + (float)(room.l_in_inch/12); br = room.b_in_feet + (float)(room.b_in_inch/12); height = room.h_in_feet + (float)(room.h_in_inch/12); double volm = (double)len * br*height; cout Customer[i].D_O_sale.day>>c>>Customer[i].D_O_sale.mon>>c>>Customer[i].D_O_sale.yr; cout Customer[i].quan; cout> Customer[i].unit_pr; cout Customer[i].discount; cout Customer[i].sales_tax; } clrscr(); cout cin>>client[i].year; coutclient[i].rate; coutclient[i].amount; x= client[i].amount*pow(( 1.0+client[i].rate/100.0), client[i].year); compound=x-client[i].amount; } for(i=0;i represented by a C structure having member/fields for the first name, last name and student number (an integer). You should sort first by first name, then by last name (this way they will be fully sorted by name when you finish). You can read the information in from the console or from a file, and should print out the sorted list to the console. You can assume some reasonable upper limit on number of records (say, 25) if you want to use an array. Ans. #include #include #include #include struct student { char fname[20]; char lname[20]; int id_no; }; typedef student S; void main() { S s[5]; char tfname[20], tlname[20]; int t; clrscr(); for(int i = 0; i strcpy(s[i].fname,s[j].fname); strcpy(s[j].fname,tfname); strcpy(tlname,s[i].lname); strcpy(s[i].lname,s[j].lname); strcpy(s[j].lname,tlname); t=s[i].id_no; s[i].id_no=s[j].id_no; s[j].id_no=t; } } } } } cout cin >> c[i].income; } char ch = 'y'; char cap[30]; int flag = 0; cin.ignore(); cout