1. write the missing statement to display 19 on the screen.
void main()
{
char x = '19';
printf("%d",x);
}
Ans: char x="\x19"
Explanation: '%x' is a format specifier to print a hexadecimal char constant. In order to write a hexadecimal char, we have to write '\x' preceding the char const. so in order to get the output 19, we have to write char x='\x19' at time of declaration.
2. find the output. void main()
{
float a=4;
int i=2;
printf("%f %d",i/a,i/a);
printf("%d %f",i/a,i/a);
}
Ans: 0.500000 0 0 0.000000.
Explanation: as a is of float datatype and i is of int datatype, so the result of i/a will definitely be a float. in the case of 1st printf(), the 1st format specifier is %f so the result gets printed in float and as the 2nd format specifier is %d, the result get printed in integer format so the putput of 1st printf() is 0.500000 and
3. Debug the program to get the output 300.
void main()
{
short int x;
x=300*300/300;
printf ("%d",x);
}
Ans: x=300*300/300;
Explanation: In order to get the output 300, we have to write either x=300*300/300 or x=300*300L/300. 300 is a short int but 300*300 result in long which can not be stored in an interger which is of 2 byte. so if we make any of the two 300 as long int by writting 300L, then the result will be a ling int in which 90000 can be stored perfectly. when an operation is done between two different data type of different size, then the shotter data type gets converted to the higher data type and the result also belongs to higher data type.
4.What is the problem in the following code.
void main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%if\n",f%g);
printf("%if\n",fmod(f,g));
}
4.compilation error
Explanation: The individual member in an enum datatype are constants and hence their value can not be changed(so ++k is illigal). the operators << and % work with integral data only. The prototype of fmod() is present in the math.h and the purpose of this function is to return the remainder where both the divisor and the devidend belong to float data type.
5.Find the output.
void main()
{
char ch=291;
printf("%d%d%c",32770,ch,ch);
}
5. Ans: -32770 0 #
Explanation:32770 is aint which will take 4 bytes to get stored in th memory. Its memory representation woll be 11111111 00000000. but in the printf() statement, the first format specifier is %hd and hence it will take 1st 2 bytes of data and print 32766 and the 2nd %hd will take the 2 bytes of data whose value is 35. As all the integral data type maintain a cycle internally in the memory, the value stored in ch will be converted to 35 which is the ASCII value of '#'.
6. Find the output.
void main()
{
int i=10,j=11,k=12;
printf("%x%x%ci%x",j,i,'s',k);
}
6. Ans: Basic
Explanation: %x is the format specifier for the hexadecimal char constant, simply int i, j and k will get converted to its equivalent hexadecimal constants which is nothing but B a and c respectively. So the final output will be Basic.
7. Find the output.
void main()
{
printf("%d\n",'-'-'-'-'/'/'/');
}
7. Ans: -1
Explanation: Anything writte within a pair of single quote(' ') means it is a char constant. All the char constant are having a perticular ASCII value. Operator '/' is having higher precedence than '-'. So, in the statement '-'-'-'-'/'/'/' 1st the division will be performed and then the substraction. Hence the output will be -1.
8.Find the output.
void main()
{
int x=10,y=5,p,q;
p=x>9;
q=x>3&&y!=3;
printf("p=%d q=%d",p,q);
}
8. Ans: 1 1
Explanation: Value of x is 10, the statement x>9 is true and hence will return 1. Also x >3 and y!=3 both are true and hence of q will be 1. So the output will be 1 1.
9.What is the problem in the following code?
float x=9.0;
main()
{
unsigned float x=8.0;
if (x=x)
printf("hello");
else
printf("Hi");
}
9. Ans: Compilation error
Explanation: unsigned modifier cannot be used with float data type; it can only be used with the integral data type.
10.find the output.
void main()
{
undigned volatile static const int i=-5;
+i;
printf("%d",i);
printf("%d",sizeof(i));
}
10. Ans: -5 4
Explanation: Here +i is a dummy statement. It has no effect in the given code.
11.Find the output.
void main()
{
double a=11.11,b;
int *ptr;
ptr=(int*)&a;
b=*ptr;
printf("%f",*ptr);
}
12.Find the output.
void main()
{
int a=6;
float b=3.000000f;
b=b*2;
if (a==b)
{
a=a+b;
printf("%d",a);
}
else
{
a=a-b;
printf("%d",a);
}
}
12.Ans:
Explanation: While comparision between the two values of different date type is done then the lower type gets converted to the higher data tyoe. So value of it to gets converted to float and then the comparison takes place. So the condition inside the if statement becomes true and hence the output is 12.
13.Find the output.
void main()
{
if (~0==(undigned int)~1)
printf("%d %u",~0,(unsigned int)~1);
else
printf("%d",~1);
}
13. Ans: -1 and 65535
Explanation: ~0 is the 1's complement of 0 which is nothing but 11111111 which is the binary representation of -1 as all the negative number are stored in the memory in the form of their 2's complements. So, the consition will be true and the output will be -1 and 65535.
14.Find the output.
void main()
{
printf("%s","c marathon"+2);
}
14. Ans: Marathon
Explanation: In printf the format specifier %s requires the base address of a string in order to print it. Here we are giving "C Marathon"+2, so the address will get incremented by 2 bytes and now the address given to the printf function is the base address of M. so the output will be equal.
15. Find the output.
int main()
{
undigned int a=-1;
int b;
b=`0;
if(a==b)
printf("equal");
else
printf("not equal");
return 0;
}
15. Ans: equal
Explanation: ~is the 1's complementing operator. So 1's complement of 0 is 11111111 which is the binary represenation of -1 as all negative numbers are stored in the form of their 2's complements. So the condition inside the of statement will be true and hence the output will be equal.
16.Find the output.
void main(){
printf("%d"+0,145);
printf("%f"+1,145);
}
16. Ans: 145f
Explanation: Inthe first case %d +0 returns the address of string %d which will print 145. In the second printf %f +1 returns the address of f. Hence f gets printed and the final putput is 145f.
17.What is the problem in the following code?
void min()
{
int const *p=5;
int q;
p=&q;
printf("%d",++(*p));
}
17. Can't modify a constant object.
Explanation: Here p is a pointer which points to a constant object. So we cannot change the value of the object by using pointer, but it can be change using the name of the object. So, when we are writting ++(*p), it will give an error that it to cannot modify a constant object.
18.What is the problem in the following code?
typedef enum error Type
{
warning,error,exception,
}error;
void main()
{
error g1;
g1=1;
printf("%d",g1);
}
18. Ans: Compilation error
Explanation: The name of the enum type must be an identifier. Here there is a space between error and type which is not an identifier. The 2nd problem is that we cannot give the same name to the enum variable and to its individual members.
19.Find the output.
void main()
{
int const i=5;
int *p;
p=&i;
printf("%d",*(p++));
}
19. Ans: 5
Explanation: Here i is a constant object whose value cannot be changed. We are storing the address of i in a pointer p. then when we *(p++), printf() will print value of *p (value at the address p i.e. 5) and then p will get incremented. so, the output will be 5.
20.Find the output.
void main()
{
int const i=5;
int const *p;
p=&i;
printf("%d",*(p++));
}
20.Ans: 5
Explanation: Here i is a constant object and also p is a constant pointer which points to a const object. The address of i is assigned to p. then we are writing *(P++). So printf() will print value of *p(i.e. 5) and then p will get incremented.
21.What is the problem in the following code?
void main()
{
int i=5;
int const *p;
p=&i;
printf ("%d",(*p)++);
}
21.Ans: Cannot modify a constant object.
Explanation: Here i is an int and p is a pointer to a constant object. (*p)++ means the value at address p is incremented, but here p is a pointer to a constant so compiler flags an error.
22.What is the problem in the following code?
void main()
{
int i=5;;;
int j=24;;;
printf("%d %d",i,j);;;;
}
22.Ans: Declaration is not allowed here.
Explanation: There is no problem in the following code in linux. But in case of Turbo Cit is a problem because, all the declarations should be done at the first in the program, before writing the other statements. Here there are two null statements before the declaration of j. so the compiler will flag an error.
23.What is the problem in the following code?
void main()
{
int i=5;
int j=24;
printf("%d %d",i,j);;
}
23. Ans: There is no problem in the given code.
Explanation: Here the declaration of i and j are done before multiple number of Null statements terminator. That's why it will show any error.
24.What is the problem in the following code?
void main()
{
void a=5;
printf("%d",a);
}
24.Ans: size of void is unknown
Explanation: void is also known as generic data type. Its size is unknown. It is not known to the compiler that how much byte it should allocate for a. So, it will flag an error.
25.Find the output.
void main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
25. Ans: ha
Explanation: the escape sequence \n is for a new line char, \b shifts the control to the last char it printed, \r shifts the control to the beginning of the same line. Hence 1st ab will be printed then it is changed to asi and lastly changed to hai.
26.What is the problem in the following code?
enum outside
{
sunday,
monday,
tuesday,
};
void main()
{
enum inside
{
wednesday,
thursday,
sunday
};
printf("%d",sunday);
}
26. Ans: 2
Explanation: We cannot declare two eenum ariable haveing same name of the individual member both inside the same scope. But, in a program, one is made global another is made local then, this can be done. As inside a scope, priority of local variable is higher then the global, so will print 2.
27.Why condition is false in the following statement?
float x=4.6;
if (x==4.6)
printf("hello");
else
printf("hi");
27. Ans: hi
Explanation: As 4.6 is recurring binary, and as the no of precision digits in float and double are different (in float it is up to 6 digits and in double it is up to 10 digits), although while comparison float will get converted to double, the value in the both will not the same. So the output will be hi.
28.What is the problem in the following code?
void main()
{
char x='\477';
printf(%c",x);
}
28. Ans: Compilation error
Explanation: When we are writting something preceded by '\' and within a pair of singlequote('\'), then it becomes an octal char constant whose maximum range is 377. Hence it will show error as numeric costant is too large.
29.What is the problem in the following code?
void main()
{
enum xxx(a=32675,b,c,d);
}
29. Ans: Numeric constant too large
Explanation: enum data type is used to create only a sequence of short integral constants whose maximum range is 32767. More than that it cannot hold . Here if a is initialized with 32765, then value of d will be 2147483647 which is more than the range of short int. So the maximum possible value of a can be 2147483647.
30.Write the missing statement so that the output will be 0 1 2. There are four such statements write all the four.
void main()
{
/*add here*/
printf("%d %d %d ",black,blue,green) ;
}
30. Ans: printf("brainq");
Explanation: We know that printf() returns the number of character it prints on the screen. So, if we will write b=printf("brainq"); then first brainq will be printed and then printf() will return 6 which will get stored in b. So the output will be brainq25 6.
31. Write the missing statement so that the output will be brainq25 6.
void main()
{
int a,b;
a=25,86,74;
b=/*add here*/
printf("%d %d",a,b);
}
31. Ans:
1 enum col(BLACK, BLUE, GREEN);
2 #include<graphics.h>
3 int BLACK=0,BLUE=1,GREEN=2;
Explanation: In Linux BLACK, BLUE, GREEN are constants, and the value 0,1,2 respectively.
1 the enum data type creates a sequence of short integral constants and by default starting value is 0 for the first one. It can also be changed by providing a value to it.
2 In Turbo C in the header file graphics.h BLACK, BLUE and GREEN are prdefined constants with value 0,1,2 respectively.
3 BLACK,BLUE and GREEEN are declared as integer variable with value 0,1,2.
32. how many time the following loop will continue?
void main()
{
short int x=1;
while(x)
{
printf("c marathon");
x++;
}
}
32. Ans: 65535
Explanation:The loop will run for 32767 times. After that zero will occur and the condition inside while() will become false
33. Write one line statement to change the byte order of an short integer,(means first byte will be second byte and second byte will be first byte).
33. Ans: x=x<<8 | x>>8;
Explanation: This statement will change the byte order of an integer. x is first shifted left 8 times and bitwise operated (or) with x shifted right 8 times.
34. Find the output.
void main()
{
char ch;
for(ch='0';ch<='\377';ch++)
printf("%d",ch);
}
34. Ans: No output
Explanation: The decimal value of '\377' is 255. But in the for loop, the value of ch (ASCII value 48) is compared with '\377', 255 will be converted to signed value, i.e.,-1. So the condirion in the for loop is not satisfied, hence no output.
35.How many times the following loop will continue?
void main()
{
short int i=1;
for (i=1;i<=40000;i++)
printf("\nHello bbsr");
}
35. Ans: Infinite loop
Explanation: As i is an int and its maximum rnge is 32767, which is always less than 40000(when it exceed 32767, enters to integer cycle), so results in infinite loop.
36.find the output.
void main()
{
int zero,zero1;
zero=10/0;
zero1=0/0;
printf("%d %d",zero,zero1);
}
36. Ans: In Linux this is runtime error, i.e., floating pointy exception But inTurbo C this is 100 because
Explanation:In Turbo C any number dicided by 0, the result will be that number and in ASII C there will be runtime error.
37.Write the source code to check if the system is little-endian or big-endian system.
37.Ans:
void main()
{
int x=300;
if ((*((unsigned char*)&x)==1)&&(*((unsigned char*)&x+1)==44))
printf("BIG ENDIAN");
else
printf(";ITTLE ENDIAN");
}
38.Design a program that will calculate the size of any variable.
38. Ans:
#define any_size(any) (char*)(&any)-(char*)((&any)-1)
void main()
{
//any type of variable like int c;
printf("%d",any_size(c));
}
39.Debug the following program.
#include"ganga.h"
void main()
{
show();
}
int const x=80;
//ganga.h
void show ()
{
extern int x++;
printf("%d",x);
}
39. Ans: Declaraction syntax error
Explanation: To debug the program in ganga.h the statement extern int x++; should be change to extern int x; The statement extern int x; tells the compiler that the variable x is defined some where else in the program or may be in other file.
40.give the memory representation of float x=20.50.
40. Ans:
float x=20.5;
The binary equivalent of 20=10100 and of 0.5= .1
That means, 20.5=10100.1
In normalized form, it will be 1.01001 *e4.
Accoding to the above example, the exponent 4 is added with bias 127and stored in exponent 8bits.
127
+ 4
131 - 10000011
41.Write the missing statement to debug the program.
41. Ans: int f(int,float);
Explanation: If the function is called before its execution its prototype is necessary before the function call.
void main()
{
int a;
a=f(10,3.14);
printf("%d",a);
}
f(int aa,float bb)
{
return((float)aa+bb);
}
42.How to check, whether cycle is present or not in integer data type?
42.Ans:
#include"limits.h"
void main()
{
if (INT_MAX+1==INT_MIN)
printf("CYCLE IS PRESEN");
else
printf("CYCLE IS NOT PRESENT");
}
43.Write an efficient algorithm to find number of hits are set to 1 in 16 bits short integer (chexking each bit using condition is not the solution).
43. Ans:
void main()
{
int n;
intc=0;
printf("\nEnter any mumber:");
scanf("%d",&n);
while(x>=1)
{
c++;
n=n&n-1;
}
printf("%d",c);
}
44.find the output.
void main()
{
const int x=get();
printf(%d),x ;
}
get()
{
return(20);
}
44.Ans: 20
Explanation: int x is a auto variable which is created and initilized at runtime. So at the runtime it will initialized with the return value of function get() which is 20.
45.This is a program to count the number of bit set to 1.But this only works for positive number. Modify the program or redesign the program to do the same for negative number also.
void main()
{
int x ;
int c=0;
printf("\nEnter any number :");
scanf("%d",&x);
while (x>=1)
{
c++;
n=n&n-1;
}
printf("%d",c);
}
45.
void main()
{
int n;
int c=0,i=0;
printf("\nEnter any number:");
scanf("%d",&n);
while(i<=31)
{
if (n&1==1)
c++;
n=n>>1;
i++;
}
printf("&d",c);
}
46.In which line of the following program it will show error.
void main()
{
void *p;
int **p1;
int a;
a=129;
p=&a;
p1=&p;
printf("*p=%d\n\n",*p);
printf("p=%d")
}
46. Ans: Linne no 9
Explanation: We can't dereference a void type of pointer.
47.Find the output.
void man()
{
int x=300;
printf("%d",(char)x/2);
}
47.Ans: 22
Explanation: 300 is first typecasted to char data type that will give 44(due to the presence of cycle in char data type) and then divide by 2 which gives 22.
48.Find the output.
void main()
{
int x=30000;
x=x+30000;
printf("%d",x);
}
48. Ans: -32536
Explanation: The maximum limit of the short integer variable is 32767. We are assigning 33000 to x which results in overflow and due to the presence of cycle in integer datatype 33000 wraps to give -32536.
49.Find the output.
void main()
{
float x=4.3;
if (x==4.3)
printf("Hello");
else
printf("Hi");
}
49.Ans: Hi
By default the floating point constant is double. As 4.3 is a recurring binary the memory representation of both x and 4.3 will differ. So, the if condition is false and the output will be Hi.
50.Find the output.
void main()
{
int a=0x1234;
a=a>>12;
a=a<<12;
printf("%x",a);
}
50.Ans: 1000
Explanation: 0x1234's memory representation is 0001 0010 0011 0100. When x is shifted 12 bits right we get 0000 0000 0000 0001.Again x is shifted 12 bit left now we get 0001 0000 0000 0000. That equals to 0x1000.
void main()
{
char x = '19';
printf("%d",x);
}
Ans: char x="\x19"
Explanation: '%x' is a format specifier to print a hexadecimal char constant. In order to write a hexadecimal char, we have to write '\x' preceding the char const. so in order to get the output 19, we have to write char x='\x19' at time of declaration.
2. find the output. void main()
{
float a=4;
int i=2;
printf("%f %d",i/a,i/a);
printf("%d %f",i/a,i/a);
}
Ans: 0.500000 0 0 0.000000.
Explanation: as a is of float datatype and i is of int datatype, so the result of i/a will definitely be a float. in the case of 1st printf(), the 1st format specifier is %f so the result gets printed in float and as the 2nd format specifier is %d, the result get printed in integer format so the putput of 1st printf() is 0.500000 and
3. Debug the program to get the output 300.
void main()
{
short int x;
x=300*300/300;
printf ("%d",x);
}
Ans: x=300*300/300;
Explanation: In order to get the output 300, we have to write either x=300*300/300 or x=300*300L/300. 300 is a short int but 300*300 result in long which can not be stored in an interger which is of 2 byte. so if we make any of the two 300 as long int by writting 300L, then the result will be a ling int in which 90000 can be stored perfectly. when an operation is done between two different data type of different size, then the shotter data type gets converted to the higher data type and the result also belongs to higher data type.
4.What is the problem in the following code.
void main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%if\n",f%g);
printf("%if\n",fmod(f,g));
}
4.compilation error
Explanation: The individual member in an enum datatype are constants and hence their value can not be changed(so ++k is illigal). the operators << and % work with integral data only. The prototype of fmod() is present in the math.h and the purpose of this function is to return the remainder where both the divisor and the devidend belong to float data type.
5.Find the output.
void main()
{
char ch=291;
printf("%d%d%c",32770,ch,ch);
}
5. Ans: -32770 0 #
Explanation:32770 is aint which will take 4 bytes to get stored in th memory. Its memory representation woll be 11111111 00000000. but in the printf() statement, the first format specifier is %hd and hence it will take 1st 2 bytes of data and print 32766 and the 2nd %hd will take the 2 bytes of data whose value is 35. As all the integral data type maintain a cycle internally in the memory, the value stored in ch will be converted to 35 which is the ASCII value of '#'.
6. Find the output.
void main()
{
int i=10,j=11,k=12;
printf("%x%x%ci%x",j,i,'s',k);
}
6. Ans: Basic
Explanation: %x is the format specifier for the hexadecimal char constant, simply int i, j and k will get converted to its equivalent hexadecimal constants which is nothing but B a and c respectively. So the final output will be Basic.
7. Find the output.
void main()
{
printf("%d\n",'-'-'-'-'/'/'/');
}
7. Ans: -1
Explanation: Anything writte within a pair of single quote(' ') means it is a char constant. All the char constant are having a perticular ASCII value. Operator '/' is having higher precedence than '-'. So, in the statement '-'-'-'-'/'/'/' 1st the division will be performed and then the substraction. Hence the output will be -1.
8.Find the output.
void main()
{
int x=10,y=5,p,q;
p=x>9;
q=x>3&&y!=3;
printf("p=%d q=%d",p,q);
}
8. Ans: 1 1
Explanation: Value of x is 10, the statement x>9 is true and hence will return 1. Also x >3 and y!=3 both are true and hence of q will be 1. So the output will be 1 1.
9.What is the problem in the following code?
float x=9.0;
main()
{
unsigned float x=8.0;
if (x=x)
printf("hello");
else
printf("Hi");
}
9. Ans: Compilation error
Explanation: unsigned modifier cannot be used with float data type; it can only be used with the integral data type.
10.find the output.
void main()
{
undigned volatile static const int i=-5;
+i;
printf("%d",i);
printf("%d",sizeof(i));
}
10. Ans: -5 4
Explanation: Here +i is a dummy statement. It has no effect in the given code.
11.Find the output.
void main()
{
double a=11.11,b;
int *ptr;
ptr=(int*)&a;
b=*ptr;
printf("%f",*ptr);
}
12.Find the output.
void main()
{
int a=6;
float b=3.000000f;
b=b*2;
if (a==b)
{
a=a+b;
printf("%d",a);
}
else
{
a=a-b;
printf("%d",a);
}
}
12.Ans:
Explanation: While comparision between the two values of different date type is done then the lower type gets converted to the higher data tyoe. So value of it to gets converted to float and then the comparison takes place. So the condition inside the if statement becomes true and hence the output is 12.
13.Find the output.
void main()
{
if (~0==(undigned int)~1)
printf("%d %u",~0,(unsigned int)~1);
else
printf("%d",~1);
}
13. Ans: -1 and 65535
Explanation: ~0 is the 1's complement of 0 which is nothing but 11111111 which is the binary representation of -1 as all the negative number are stored in the memory in the form of their 2's complements. So, the consition will be true and the output will be -1 and 65535.
14.Find the output.
void main()
{
printf("%s","c marathon"+2);
}
14. Ans: Marathon
Explanation: In printf the format specifier %s requires the base address of a string in order to print it. Here we are giving "C Marathon"+2, so the address will get incremented by 2 bytes and now the address given to the printf function is the base address of M. so the output will be equal.
15. Find the output.
int main()
{
undigned int a=-1;
int b;
b=`0;
if(a==b)
printf("equal");
else
printf("not equal");
return 0;
}
15. Ans: equal
Explanation: ~is the 1's complementing operator. So 1's complement of 0 is 11111111 which is the binary represenation of -1 as all negative numbers are stored in the form of their 2's complements. So the condition inside the of statement will be true and hence the output will be equal.
16.Find the output.
void main(){
printf("%d"+0,145);
printf("%f"+1,145);
}
16. Ans: 145f
Explanation: Inthe first case %d +0 returns the address of string %d which will print 145. In the second printf %f +1 returns the address of f. Hence f gets printed and the final putput is 145f.
17.What is the problem in the following code?
void min()
{
int const *p=5;
int q;
p=&q;
printf("%d",++(*p));
}
17. Can't modify a constant object.
Explanation: Here p is a pointer which points to a constant object. So we cannot change the value of the object by using pointer, but it can be change using the name of the object. So, when we are writting ++(*p), it will give an error that it to cannot modify a constant object.
18.What is the problem in the following code?
typedef enum error Type
{
warning,error,exception,
}error;
void main()
{
error g1;
g1=1;
printf("%d",g1);
}
18. Ans: Compilation error
Explanation: The name of the enum type must be an identifier. Here there is a space between error and type which is not an identifier. The 2nd problem is that we cannot give the same name to the enum variable and to its individual members.
19.Find the output.
void main()
{
int const i=5;
int *p;
p=&i;
printf("%d",*(p++));
}
19. Ans: 5
Explanation: Here i is a constant object whose value cannot be changed. We are storing the address of i in a pointer p. then when we *(p++), printf() will print value of *p (value at the address p i.e. 5) and then p will get incremented. so, the output will be 5.
20.Find the output.
void main()
{
int const i=5;
int const *p;
p=&i;
printf("%d",*(p++));
}
20.Ans: 5
Explanation: Here i is a constant object and also p is a constant pointer which points to a const object. The address of i is assigned to p. then we are writing *(P++). So printf() will print value of *p(i.e. 5) and then p will get incremented.
21.What is the problem in the following code?
void main()
{
int i=5;
int const *p;
p=&i;
printf ("%d",(*p)++);
}
21.Ans: Cannot modify a constant object.
Explanation: Here i is an int and p is a pointer to a constant object. (*p)++ means the value at address p is incremented, but here p is a pointer to a constant so compiler flags an error.
22.What is the problem in the following code?
void main()
{
int i=5;;;
int j=24;;;
printf("%d %d",i,j);;;;
}
22.Ans: Declaration is not allowed here.
Explanation: There is no problem in the following code in linux. But in case of Turbo Cit is a problem because, all the declarations should be done at the first in the program, before writing the other statements. Here there are two null statements before the declaration of j. so the compiler will flag an error.
23.What is the problem in the following code?
void main()
{
int i=5;
int j=24;
printf("%d %d",i,j);;
}
23. Ans: There is no problem in the given code.
Explanation: Here the declaration of i and j are done before multiple number of Null statements terminator. That's why it will show any error.
24.What is the problem in the following code?
void main()
{
void a=5;
printf("%d",a);
}
24.Ans: size of void is unknown
Explanation: void is also known as generic data type. Its size is unknown. It is not known to the compiler that how much byte it should allocate for a. So, it will flag an error.
25.Find the output.
void main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
25. Ans: ha
Explanation: the escape sequence \n is for a new line char, \b shifts the control to the last char it printed, \r shifts the control to the beginning of the same line. Hence 1st ab will be printed then it is changed to asi and lastly changed to hai.
26.What is the problem in the following code?
enum outside
{
sunday,
monday,
tuesday,
};
void main()
{
enum inside
{
wednesday,
thursday,
sunday
};
printf("%d",sunday);
}
26. Ans: 2
Explanation: We cannot declare two eenum ariable haveing same name of the individual member both inside the same scope. But, in a program, one is made global another is made local then, this can be done. As inside a scope, priority of local variable is higher then the global, so will print 2.
27.Why condition is false in the following statement?
float x=4.6;
if (x==4.6)
printf("hello");
else
printf("hi");
27. Ans: hi
Explanation: As 4.6 is recurring binary, and as the no of precision digits in float and double are different (in float it is up to 6 digits and in double it is up to 10 digits), although while comparison float will get converted to double, the value in the both will not the same. So the output will be hi.
28.What is the problem in the following code?
void main()
{
char x='\477';
printf(%c",x);
}
28. Ans: Compilation error
Explanation: When we are writting something preceded by '\' and within a pair of singlequote('\'), then it becomes an octal char constant whose maximum range is 377. Hence it will show error as numeric costant is too large.
29.What is the problem in the following code?
void main()
{
enum xxx(a=32675,b,c,d);
}
29. Ans: Numeric constant too large
Explanation: enum data type is used to create only a sequence of short integral constants whose maximum range is 32767. More than that it cannot hold . Here if a is initialized with 32765, then value of d will be 2147483647 which is more than the range of short int. So the maximum possible value of a can be 2147483647.
30.Write the missing statement so that the output will be 0 1 2. There are four such statements write all the four.
void main()
{
/*add here*/
printf("%d %d %d ",black,blue,green) ;
}
30. Ans: printf("brainq");
Explanation: We know that printf() returns the number of character it prints on the screen. So, if we will write b=printf("brainq"); then first brainq will be printed and then printf() will return 6 which will get stored in b. So the output will be brainq25 6.
31. Write the missing statement so that the output will be brainq25 6.
void main()
{
int a,b;
a=25,86,74;
b=/*add here*/
printf("%d %d",a,b);
}
31. Ans:
1 enum col(BLACK, BLUE, GREEN);
2 #include<graphics.h>
3 int BLACK=0,BLUE=1,GREEN=2;
Explanation: In Linux BLACK, BLUE, GREEN are constants, and the value 0,1,2 respectively.
1 the enum data type creates a sequence of short integral constants and by default starting value is 0 for the first one. It can also be changed by providing a value to it.
2 In Turbo C in the header file graphics.h BLACK, BLUE and GREEN are prdefined constants with value 0,1,2 respectively.
3 BLACK,BLUE and GREEEN are declared as integer variable with value 0,1,2.
32. how many time the following loop will continue?
void main()
{
short int x=1;
while(x)
{
printf("c marathon");
x++;
}
}
32. Ans: 65535
Explanation:The loop will run for 32767 times. After that zero will occur and the condition inside while() will become false
33. Write one line statement to change the byte order of an short integer,(means first byte will be second byte and second byte will be first byte).
33. Ans: x=x<<8 | x>>8;
Explanation: This statement will change the byte order of an integer. x is first shifted left 8 times and bitwise operated (or) with x shifted right 8 times.
34. Find the output.
void main()
{
char ch;
for(ch='0';ch<='\377';ch++)
printf("%d",ch);
}
34. Ans: No output
Explanation: The decimal value of '\377' is 255. But in the for loop, the value of ch (ASCII value 48) is compared with '\377', 255 will be converted to signed value, i.e.,-1. So the condirion in the for loop is not satisfied, hence no output.
35.How many times the following loop will continue?
void main()
{
short int i=1;
for (i=1;i<=40000;i++)
printf("\nHello bbsr");
}
35. Ans: Infinite loop
Explanation: As i is an int and its maximum rnge is 32767, which is always less than 40000(when it exceed 32767, enters to integer cycle), so results in infinite loop.
36.find the output.
void main()
{
int zero,zero1;
zero=10/0;
zero1=0/0;
printf("%d %d",zero,zero1);
}
36. Ans: In Linux this is runtime error, i.e., floating pointy exception But inTurbo C this is 100 because
Explanation:In Turbo C any number dicided by 0, the result will be that number and in ASII C there will be runtime error.
37.Write the source code to check if the system is little-endian or big-endian system.
37.Ans:
void main()
{
int x=300;
if ((*((unsigned char*)&x)==1)&&(*((unsigned char*)&x+1)==44))
printf("BIG ENDIAN");
else
printf(";ITTLE ENDIAN");
}
38.Design a program that will calculate the size of any variable.
38. Ans:
#define any_size(any) (char*)(&any)-(char*)((&any)-1)
void main()
{
//any type of variable like int c;
printf("%d",any_size(c));
}
39.Debug the following program.
#include"ganga.h"
void main()
{
show();
}
int const x=80;
//ganga.h
void show ()
{
extern int x++;
printf("%d",x);
}
39. Ans: Declaraction syntax error
Explanation: To debug the program in ganga.h the statement extern int x++; should be change to extern int x; The statement extern int x; tells the compiler that the variable x is defined some where else in the program or may be in other file.
40.give the memory representation of float x=20.50.
40. Ans:
float x=20.5;
The binary equivalent of 20=10100 and of 0.5= .1
That means, 20.5=10100.1
In normalized form, it will be 1.01001 *e4.
Accoding to the above example, the exponent 4 is added with bias 127and stored in exponent 8bits.
127
+ 4
131 - 10000011
41.Write the missing statement to debug the program.
41. Ans: int f(int,float);
Explanation: If the function is called before its execution its prototype is necessary before the function call.
void main()
{
int a;
a=f(10,3.14);
printf("%d",a);
}
f(int aa,float bb)
{
return((float)aa+bb);
}
42.How to check, whether cycle is present or not in integer data type?
42.Ans:
#include"limits.h"
void main()
{
if (INT_MAX+1==INT_MIN)
printf("CYCLE IS PRESEN");
else
printf("CYCLE IS NOT PRESENT");
}
43.Write an efficient algorithm to find number of hits are set to 1 in 16 bits short integer (chexking each bit using condition is not the solution).
43. Ans:
void main()
{
int n;
intc=0;
printf("\nEnter any mumber:");
scanf("%d",&n);
while(x>=1)
{
c++;
n=n&n-1;
}
printf("%d",c);
}
44.find the output.
void main()
{
const int x=get();
printf(%d),x ;
}
get()
{
return(20);
}
44.Ans: 20
Explanation: int x is a auto variable which is created and initilized at runtime. So at the runtime it will initialized with the return value of function get() which is 20.
45.This is a program to count the number of bit set to 1.But this only works for positive number. Modify the program or redesign the program to do the same for negative number also.
void main()
{
int x ;
int c=0;
printf("\nEnter any number :");
scanf("%d",&x);
while (x>=1)
{
c++;
n=n&n-1;
}
printf("%d",c);
}
45.
void main()
{
int n;
int c=0,i=0;
printf("\nEnter any number:");
scanf("%d",&n);
while(i<=31)
{
if (n&1==1)
c++;
n=n>>1;
i++;
}
printf("&d",c);
}
46.In which line of the following program it will show error.
void main()
{
void *p;
int **p1;
int a;
a=129;
p=&a;
p1=&p;
printf("*p=%d\n\n",*p);
printf("p=%d")
}
46. Ans: Linne no 9
Explanation: We can't dereference a void type of pointer.
47.Find the output.
void man()
{
int x=300;
printf("%d",(char)x/2);
}
47.Ans: 22
Explanation: 300 is first typecasted to char data type that will give 44(due to the presence of cycle in char data type) and then divide by 2 which gives 22.
48.Find the output.
void main()
{
int x=30000;
x=x+30000;
printf("%d",x);
}
48. Ans: -32536
Explanation: The maximum limit of the short integer variable is 32767. We are assigning 33000 to x which results in overflow and due to the presence of cycle in integer datatype 33000 wraps to give -32536.
49.Find the output.
void main()
{
float x=4.3;
if (x==4.3)
printf("Hello");
else
printf("Hi");
}
49.Ans: Hi
By default the floating point constant is double. As 4.3 is a recurring binary the memory representation of both x and 4.3 will differ. So, the if condition is false and the output will be Hi.
50.Find the output.
void main()
{
int a=0x1234;
a=a>>12;
a=a<<12;
printf("%x",a);
}
50.Ans: 1000
Explanation: 0x1234's memory representation is 0001 0010 0011 0100. When x is shifted 12 bits right we get 0000 0000 0000 0001.Again x is shifted 12 bit left now we get 0001 0000 0000 0000. That equals to 0x1000.
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