Answers to Practice Problems

Transcription

1 Answers to Practice Problems Prepared by:dilip Kumar Gangwar(Faculty,CS/IT DEPTT GEHU) 1)C Language is developed by- 1.Bjarne stroutstrup 2.Dennis Ritchie 3.R.Byrant 4.None of the above 2)C is a a)middle level language b)high level c)low level d)c is not a language 3)Header files have the extension a).c b).hf c).txt d).h 4)Header files have a)function declaration,but not the function definition b)function definition not function declaration d)none of these 5)printf() and scanf() are

2 a)function and datatype respectively b) datatype and function respectively c)both functions d)data type and datatypes respectively 6)main() is a a) function b)datatype c)variable d)header file 7)stdio.h is a a)header file b)datatype c)variable 8)stdio.h has declaration of a)getch() and clrscr() b)printf() and scanf() c)scanf() and getch() d)only getch() 9)conio.h has declaration of a)getch() and clrscr() b)printf() and scanf() c)scanf() and getch() d)only getch()

3 10)return statement can be used to a)simply return the control to a function from where it has come without specifying any value b) return the control to OS c)return value to a function function from where it has come d)return to home. e)both (A) and (c) 11) The statement int a is a a) definition of a b)declaration of a c)both (a) and (b) d)none of these 12) The statement int a=10 is a a) definition of variable a b)declaration of a c)initialisation of a d) Both (a) and (b),(c) 13) int main() int a=10; int a=12; printf( %d,a);

4 Ouput of it is---- a)10 12 b)12 10 c)10 d)12 e)compile time error redefinition of a 14)getch() is used to a)read or take the value from keyboard/user b)hold the output screen c)print anything on screen d)none of these 15)scanf() is used to a)read or take the value from keyboard/user b)hold the screen c)print anything on screen d)none of these 16)printf() is used to a)read or take the value from keyboard/user b)hold the screen c)print anything message and printing the values of variables on the screen d)none of these 17)clrscr() is used to

5 a)read or take the value from keyboard/user b)hold the screen c)print anything on screen d)clear the previous output on the screen 18)Full form ofide a)integrated disk environment b) Integrated development environment c)integrated datum enum 19)IDE is a)virus b)hardware which helps to print our data c)a software which provides the facility/environment to a programmar to develop programs in a programming language d) a software which provides the facility/environment to a programmar to crash the OS 20)Ascii value of semicolon ( ; ) and blank/space is a)59 32 b) what is the output of the following code..? Int main() Char s= c ; printf( %d%d\n,sizeof(s),sizeof( c )); a. 1 1 b. 1 8 c )int emp id=341;

6 Here variable name emp id is valid or not?give reason It is not valid as emp id contains space in between and a variable/identifier name can not have space/blank,comma,any special character (e.g.?,#,* ) in between them.variable name can have Underscore in between(e.g. total_salary is valid,but total salary is not valid). 23)Compiler like Turbo c/c++ or Dev c++ is a a)software b)hardware c)name of a datatype d)all of the above 24) What will be the output? int main() int a=10; int b=5; int c=5; int d=2; int e=a*d +b/c; printf("e=%d",e); 1. error none e=10*2+5/2; =20+5/2; precedence of * and / is higher than +,so check for * and /,but * and / have same priority.now check for associativity.associativity is From Left to right.so * will be evaluated first then / e= 20+2; now evaluate + e=22; 25) int main() float avg; int a=10; int b=5;

7 int c=2; avg=a+b+c/3; printf("avg=%f",avg); output a)5.66 b) precedence of / is more than + c)5 d)error avg=10+5+2/3 avg= avg=15 now typecasting will take place as = avg is float(4 bytes)=15 is integer(2 byte) //promotion so 15 has to go in avg (float) and will adjust there as float i.e implicit(conversion in assignment) will take place. And 15 will become so avg = ) int main() float avg; int a=10; int b=5; int c=2; avg=(a+b+c)/3; printf("avg=%f",avg); output

8 a)5.66 b)5.0 since ( ) has highest precedence/priority than any operator c)5 d)error avg=(10+5+2)/3 avg=(15+2)/3 avg=17/3 avg=5 again typecasting will take place float = int //implicit typecasting(conversion in assignment) avg= //promotion of 5 (int) to float 27)int main() float avg; float a=10; int b=5; int c=2; avg=a+b+c/3; printf("avg=%f",avg); output---- a)5.66 b)5.0 c) d)error float a= ; //here a will have typecasting).10 has to adjust as float a= as a is float so float=int (implicit

9 avg= /3 avg= avg= )int main() float avg; float a=10; int b=5; int c=2; avg=(a+b+c)/3; printf("avg=%f",avg); output---- a) b)5.0 c)5 d)error avg=( )/3 avg=( )/3 avg= /3 now float/int=float avg= )int main() float avg;

10 float a=10; int b=5; int c=2; avg=(a+b+c)/3; printf("avg=%d",avg); output a)5.66 b)5.0 c)5 d) or abnormal program termination printf("avg=%d",avg); //due to %d error happens as avg is float and we are printing it as integer.we cannot do typecasting in printf 30) int main() Int x,y,z; x=20++; y=++x++; z=(x+y)--; printf( %d %d %d,x,y,z); Output

11 A b) c)error L value required 20 is a constant value we cannot increment it, same for ++x++ 31) main() int a ; printf( %d,a); 1)0 2)9.88 c)a d)garbage value 32)size of a datatype is a)machine independent b)machine dependent c)always fixed 33) main() printf( %x,11);

12 a)b or B b)14 c)x d)error printf( %x,11); // x is to print hexadecimal form of a number 34) main() printf( %o,11); a)13 b)11 c)x d)error printf( %o,11); // o is to print octal form of a number 35)Comments in c are of a)2 types single line // and multiline /* */ b)4 type c)1 type

13 36)Computer understands language of a)english b) 0 s and 1 s c)assembly language d)it is fool 37)1 byte= bits a)5 b)8 c)9 d)16 38)1 kb= byte a)2^10 b)2^20 c)2^30 39)1 MB= byte a)2^10 b)2^20 c)2^30 40)1 GB= byte a)2^10 b)2^20 c)2^30 41) main()

14 char ch='a'; printf("%d %c",ch,ch); a)66 A b)65 A c)97 A d)error 42) main() char ch=130; printf("%d",ch); a)-129 b)-126 //as range of char datatype which is by default signed char is -128 to +127 so overflow takes place c)-130 d)error 43) void main() float a= ; printf("%f ",a);

15 printf("%.2f ",a); printf("%.3f",a); a) b) ) void main() float a1,b1,a2,b2,a3,b3; a1=2; b1=6.8; a2=4.2; b2=3.57; a3=9.82; b3=85.673; printf("%3.1f %4.2f\n",a1,b1); printf("%5.1f %6.2f\n",a1,b1); printf("%7.1f %8.2f\n",a1,b1); a) b) ) main() int a; int i=2; a=i++;

16 printf("%d %d",i++,a); printf( %d,i); a)3 2 4 b) ) ) ) main() int a; int i=2; a=++i; printf("%d %d ",i++,a); printf("%d,i); a)3 2 4 b) ) )2 3 3

17 47) main() int a; int i=2; a=++i + ++i; printf("%d %d",i++,a); a)4 7 b)8 4 c)4 8 d) )main() int x=4,y=3,z; z=x-- - y; //line 4 printf("%d %d %d",x,y,z); a)3 3 2 b)4 3 1 c)3 3 0 d)3 3 1

18 49) If line 4 is changed to z=x-- - y the ouput is a)3 3 2 b)4 3 1 c)3 3 0 d) )A c variable cannot contain a)blank spaces b)hyphen 3)decimal point d)allof the above 51)int a=30* evaluates to (in turbo c/c++) a) //rage of int= to b)32768 c)10000 d)0 52) main() int i=4,j=-1,k=0,y,z; y=i + 5 && j +1 k+2; printf("%d %d ",i,j); z=i+5 j+1 && k+2; printf("%d %d",y,z);

19 a) b) c) d) ) main() int i=-1,j=1,k,l; k=!i &&!j; l=!i j; printf("%d %d %d %d",i,j,k,l); a) b) c) d) ) int main() int a=10; printf("%d %d %d",a,a++,++a); a) b) c) ) int main() int x=25,y=25,z; z=x==25&&y<++x;

20 printf("\n%d %d %d ",x,y,z); a) ) int main() int x=25,y=10,z=10; x=y==z; printf("\n%d %d %d ",x,y,z); a) b) c) d) Error 57) int main() int x=100; printf("%d",20+ x++); printf("%d",20+ ++x); a) b) c) ) int main() int a=7,b=12,c=13,i; i=a b&c; printf("%d\n",i); a)13 b)15 c)garbage 59) void main() int a=5,b=3; b=++a + a b;

21 b=--b + --a *b++; printf("%d %d",a,b); a)6 79 b)6 92 c) )int c= 4*12-7>>2; a)45 b)20 c)10 d)none C=48-7>>2 C=41>>2 C=10 (41/2 2 =10.5) 61) What will be output of the following program? #include<stdio.h> int main() int a=2,b=7,c=10; c=a==b; printf("%d",c); a)10 b)0 c)7 d) 2 C=a==b Precedence of == is greater than = C=10==7 C=0 62) What will be output of the following program?give reason for your answer #include<stdio.h> void main() int x; x=10,20,30; printf("%d",x); a)10 b)20 c)30 d) e)error 63) What will be output of the following program On turbo c++(tc)? #include<stdio.h> int main() printf("%d %d %d",sizeof(3.14),sizeof(3.14f),sizeof(3.14l));

22 a) b) c) on turbo c cpmpiler: floating point constant are treated as double so 3.4 is double type,so size=8 3.2f denotes floating point constant/number size of float=4 3.2L denotes long double,so size of long double=10 On dev c++ Answer= ) What will be output of the following program? #include<stdio.h> int main() int x=10,y=20,z=5; printf("%d %d %d"); a) b)garbage answers c) d)compile time error 65) What will be output of the following program in Turbo C? #include<stdio.h> int main() int a; a=sizeof(!5.6); printf("%d",a); a)4 b)2 c)0 d)1 on Turbo C: a=sizeof(!5.6) (!non zero number)= 0 a=sizeof(0) a=2 ( as 0 is integer number ) on dev c++ a=4 (as on dev c++ size of integer=4 bytes) 66) What will be output of the following c program? #include<stdio.h> int main() long int 1a=51; //variable cannot start with digit printf("%ld",1a); a)51 b) 4 c) compilation error 67) #include<stdio.h> int main() int _=5; int =10; int ; =_+ ;

23 printf("%d", ); a)error b) 15 c)0 5)5 _ is valid identifier 68)What is the valid C statement to print the message My salary increased to 10%!? a)printf( My salary increased to 10%! ); b)printf( My salary increased to 10%%! ); c)printf( My salary increased to 10\%\! ); d)printf( My salary increased to 10/%//! ); 69) int main() int a=2; int b; b=a>7 +4 *8 && (a=a+3); printf("%d %d",a,b); a)5 0 b) 2 1 c) 2 0 d)5 1 b=a>7 +4 *8 && (a=a+3); although precedence of ( ) is highest but when there is && operator or operator in expression,then compiler calculates the lhs expressions of && or for some optimization purposes like short circuiting. so, b=a>7+28&&(a=a+3); b=a>35&&(a=a+3) b=2>35&&(a=a+3) b=0&&(a=a+3) now due to short circuit in && rhs expression of && is not evaluated b=0; and value of a has not changed a=2 70)void main() float a; a=(float)(sizeof( a )); printf( %f,a); a) b) c) on turbo c: a=(float)(sizeof( a )) character constant =integer a=(float)(2) a= on dev c++ a=float(4) size of integer=4 bytes a= )void main() int a=3,b=78,c;

24 c=printf( %d %d,a,b); printf( %d\n,c); a) b) c) d)error 71) void main() int a,b,c; c=scanf( %d%d,&a,&b); printf( %d\n,c); Suppose in scanf, input is= 5 9 a)3 b)2 c)5 2 9 d)error on success,scanf() returns no of successful characters read otherwise returns 0 here scanf() reads 2 characters/values in a and b,this answer 2 is stored in c 72) void main() int c=printf( language ); printf( %d,c); a)0 b)1 c) 8 d)2 printf() returns no of character written on success otherwise return s 0 here printf( ) writes 8 characters of language,so it returns 8,which is stored in c