Answer and explanation of questions are based
on turbo c 3.0 compilers. Answer and explanation may vary in other compilers.
1.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
char arr[7]= "Network" ;
printf( "%s" ,arr);
}
Choose all that apply:
Explanation:
Size of a character array should
one greater than total number of characters in any string which it stores. In c
every string has one terminating null character. This represents end of the
string.
So in the string “Network” , there
are 8 characters and they are ‘N’,’e’,’t’,’w’,’o’,’r’,’k’ and ‘\0’. Size of
array arr is seven. So array arr will store only first sevent characters and it
will note store null character.
As we know %s in prinf statement
prints stream of characters until it doesn’t get first null character. Since
array arr has not stored any null character so it will print garbage value.
2.
WWhat will be output when you will execute following c code?
#include <stdio.h>
void main(){
char arr[11]= "The African Queen" ;
printf( "%s" ,arr);
}
Choose all that apply: Explanation:
Size of any character array cannot
be less than the number of characters in any string which it has assigned. Size
of an array can be equal (excluding null character) or greater than but never
less than.
3.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
int const SIZE=5;
int expr;
double value[SIZE]={2.0,4.0,6.0,8.0,10.0};
expr=1|2|3|4;
printf( "%f" ,value[expr]);
}
Choose all that apply:
Explanation:
Size of any array in c cannot be
constantan variable.
4.
What will be output when you will execute following c code?
#include <stdio.h>
enum power{
Dalai ,
Vladimir =3,
Barack ,
Hillary
};
void main(){
float leader[ Dalai + Hillary ]={1.f,2.f,3.f,4.f,5.f};
enum power p= Barack ;
printf( "%0.f" ,leader[p>>1+1]);
}
Choose all that apply:
Explanation:
Size of an array can be enum
constantan.
Value of enum constant Barack will
equal to Vladimir + 1 = 3 +1 = 4
So, value of enum variable p = 4
leader[p >> 1 +1]
= leader[4 >> 1+1]
=leader[4 >> 2] //+ operator enjoy higher precedence than
>> operator.
=leader[1] //4>>2 = (4 / (2^2) = 4/4 = 1
=2
5.
What will be output when you will execute following c code?
#include <stdio.h>
#define var 3
void main(){
char *cricket[var+~0]={ "clarke" , "kallis" };
char *ptr=cricket[1+~0];
printf( "%c" ,*++ptr);
}
Choose all that apply:
Explanation:
In the expression of size of an
array can have micro constant.
var +~0 = 3 + ~0 = 3 + (-1) = 2
Let’s assume string “clarke” and
“kallis” has stored at memory address 100 and 500 respectively as shown in the following
figure:
For string “clarke”:
For string “kallis”:
In this program cricket is array of
character’s pointer of size 2. So array cricket will keep the memory address of
first character of both strings i.e. content of array cricket is:
cricket[2] = {100,500}
ptr is character pointer which is
pointing to the fist element of array cricket. So, ptr = 100
Now consider on *++ptr
Since ptr = 100 so after ++ptr ,
ptr = 101
*(++ptr) = *(101) = content of
memory address 101. From above figure it is clear that character is l.
6.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
char data[2][3][2]={0,1,2,3,4,5,6,7,8,9,10,11};
printf( "%o" ,data[0][2][1]);
}
Choose all that apply:
Explanation: %o in printf statement is used to print number in the octal format.
7.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
short num[3][2]={3,6,9,12,15,18};
printf( "%d %d" ,*(num+1)[1],**(num+2));
}
Choose all that apply:
Explanation:
*(num+1)[1]
=*(*((num+1)+1))
=*(*(num+2))
=*(num[2])
=num[2][0]
=15
And
**(num+2)
=*(num[2]+0)
=num[2][0]
=15
8.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
char *ptr= "cquestionbank" ;
printf( "%d" ,-3[ptr]);
}
Choose all that apply:
Explanation:
-3[ptr]
=-*(3+ptr)
=-*(ptr+3)
=-ptr[3]
=-103 //ASCII value of character ‘e’ is 103
9.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
long myarr[2][4]={0l,1l,2l,3l,4l,5l,6l,7l};
printf( "%ld\t" ,myarr[1][2]);
printf( "%ld%ld\t" ,*(myarr[1]+3),3[myarr[1]]);
printf( "%ld%ld%ld\t" ,*(*(myarr+1)+2),*(1[myarr]+2),3[1[myarr]]);
}
Choose all that apply:
Explanation:
Think yourself.
10.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
int array[2][3]={5,10,15,20,25,30};
int (*ptr)[2][3]=&array;
printf( "%d\t" ,***ptr);
printf( "%d\t" ,***(ptr+1));
printf( "%d\t" ,**(*ptr+1));
printf( "%d\t" ,*(*(*ptr+1)+2));
}
Choose all that apply: Explanation:
ptr is pointer to two dimension
array.
***ptr
=***&array //ptr = &array
=**array //* and & always cancel
to each other
=*arr[0] // *array = *(array +0) = array[0]
=array[0][0]
= 5
Rests think yourself.
11.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
static int a=2,b=4,c=8;
static int *arr1[2]={&a,&b};
static int *arr2[2]={&b,&c};
int * (*arr[2])[2]={&arr1,&arr2};
printf( "%d %d\t" ,*(*arr[0])[1], *(*(**(arr+1)+1)));
}
Choose all that apply:
Explanation:
Consider on the following
expression:
*(*arr[0])[1]
=*(*&arr1)[1] //arr[0] = &arr1
=*arr1[1] //* and & always cancel to each other
=*&b
=b
=4
Consider
on following expression:
*(*(**(arr+1)+1))
=
*(*(*arr[1]+1)) //*(arr+1) = arr[1]
=
*(*(*&arr2+1)) //arr[1] = &arr2
=*(*(arr2+1))
//*&arr2 = arr2
=*(arr2[1]) //*(arr2+1) = arr2[1]
= *&c //arr2[1] = &c
=
c
= 8
12.
What will be output when you will execute following c code?
#include <stdio.h>
#include <math.h>
double myfun( double );
void main(){
double (*array[3])( double );
array[0]=exp;
array[1]=sqrt;
array[2]=myfun;
printf( "%.1f\t" ,(*array)((*array[2])((**(array+1))(4))));
}
double myfun( double d){
d-=1;
return d;
}
Choose all that apply:
Explanation:
array is array of pointer to such
function which parameter is double type data and return type is double.
Consider on following expression:
(*array)((*array[2])((**(array+1))(4)))
=
(*array)((*array[2])((*array[1])(4)))
//*(array+1)
= array[1]
=
(*array)((*array[2])(sqrt(4))))
//array[1]
= address of sqrt function
=
(*array)((*array[2])(2.000000)))
=
(*array)(myfun(2.000000)))
//
array[2] = address of myfunc function
=(*array)(1.000000)
=array[0](1.000000)
=exp(1.000000)
13.
What will be output when you will execute following c code?
#include <stdio.h>
typedef struct {
char * name ;
double salary ;
} job ;
void main(){
static job a={ "TCS" ,15000.0};
static job b={ "IBM" ,25000.0};
static job c={ "Google" ,35000.0};
int x=5;
job * arr[3]={&a,&b,&c};
printf( "%s %f\t" ,(3,x>>5-4)[*arr]);
}
double myfun( double d){
d-=1;
return d;
}
Choose all that apply:
Explanation:
(3,5>>5-4)[*arr]
=(3,5>>5-4)[*arr] //x=5
= (3,5>>1)[*arr] //- operator enjoy higher
precedence than >>
= (3,2)[*arr]
//5>>1 = 5/(2^1) = 5 /2 = 2
= 2[*arr] //In c
comma is also operator.
= *(2 + *arr)
= *(*arr + 2)
=*arr[2]
=*(&c) //arr[2] = &c
=c // * and & always cancel to each other.
So,
printf("%s %f\t",c);
=> printf("%s %f\t", "Google",35000.0);
14.
What will be output when you will execute following c code?
#include <stdio.h>
union group{
char xarr [2][2];
char yarr [4];
};
void main(){
union group x={ 'A' , 'B' , 'C' , 'D' };
printf( "%c" ,x. xarr [x. yarr [2]-67][x. yarr [3]-67]);
}
Choose all that apply:
Explanation:
In
union all member variables share common memory space.
So
union member variable, array xarray will look like:
{
{‘A’,’B’},
{‘C’,’D’}
}
And
union member variable, array yarray will look like:
{
{‘A’,’B’,’C’,’D’}
}
x.xarr[x.yarr[2]-67][x.yarr[3]-67]
= x.xarr[‘C’-67][‘D’-67]
= x.xarr[67-67][68-67]
//ASCII value of ‘C’ is 67 and ‘D’ is 68.
x.xarr[0][1]
=’B’
15.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
int a=5,b=10,c=15;
int *arr[3]={&a,&b,&c};
printf( "%d" ,*arr[*arr[1]-8]);
}
Choose all that apply:
Explanation:
Member
of an array cannot be address of auto variable because array gets memory at
load time while auto variable gets memory at run time.
16.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
int arr[][3]={{1,2},{3,4,5},{5}};
printf( "%d %d %d" , sizeof (arr),arr[0][2],arr[1][2]);
}
Choose all that apply:
Explanation:
If
we will not write size of first member of any array at the time of declaration
then size of the first dimension is max elements in the initialization of array
of that dimension.
So,
size of first dimension in above question is 3.
So
size of array = (size of int) * (total number of elements) = 2 *(3*3) = 18
Default
initial value of rest elements are zero.
So above array will look like:
{
{1,2,0}
{3,4,5},
{5,0,0}
}
17.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
int xxx[10]={5};
printf( "%d %d" ,xxx[1],xxx[9]);
}
Choose all that apply:
Explanation:
If
we initialize any array at the time of declaration the compiler will treat such
array as static variable and its default value of uninitialized member is zero.
18.
What will be output when you will execute following c code?
#include <stdio.h>
#define WWW -1
enum { cat , rat };
void main(){
int Dhoni[]={2, 'b' ,0x3,01001, '\x1d' , '\111' , rat ,WWW};
int i;
for (i=0;i<8;i++)
printf( " %d" ,Dhoni[i]);
}
Choose all that apply:
Explanation:
Dhoni[0]=2
Dhoni[1]=’b’
=98 //ASCII value of character ‘b’ is
98.
Dhoni[2]= 0x3
= 3 //0x represents hexadecimal number. Decimal
value of hexadecimal 3 is also 3.
Dhoni[3]=01001
= 513 //Number begins with 0 represents octal number.
Dhoni[4] = ‘\x1d’ = 29 //’\x1d’ is hexadecimal
character constant.
Dhoni[5]
= ‘\111’ = 73 //’\111’ is octal character constant.
Dhoni[6]
=rat = 1 //rat is enum constant
Dhoni[7]
= WWW = -1 //WWW is macro constant.
19.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
long double a;
signed char b;
int arr[ sizeof (!a+b)];
printf( "%d" , sizeof (arr));
}
Choose all that apply:
Explanation:
Size of data type in TURBO C 3.0
compiler is:
S.N.
Data type
Size(In
byte)
1
char
1
2
int
2
3
double
8
Consider on the expression: !a + b
! Operator always return zero if a
is non-zero number other wisie 1. In general we can say ! operator always returns
an int type number. So
!a +b
=! (Any double type number) + Any
character type number
= Any integer type number + any
character type number
= Any integer type number
Note: In any expression lower type
data is always automatically type casted into the higher data type. In this
case char data type is automatically type casted into the int type data.
So sizeof (!a +b) = sizeof(Any int
type number) = 2
So size of array arr is 2 and its
data type is int. So
sizeof(arr) = size of array *
sizeof its data type = 2* 2= 4
20.
What will be output when you will execute following c code?
#include <stdio.h>
void main(){
char array[]= "Ashfaq \0 Kayani" ;
char *str= "Ashfaq \0 Kayani" ;
printf( "%s %c\n" ,array,array[2]);
printf( "%s %c\n" ,str,str[2]);
printf( "%d %d\n" , sizeof (array), sizeof (str));
}
Choose all that apply:
Explanation:
A
character array keeps the each element of an assigned array but a character
pointer always keeps the memory address of first element.
As
we know %s in prints the characters of stream until it doesn’t any null character
(‘\0’). So first and second printf
function will print same thing in the above program. But size of array is total numbers of its
elements i.e. 16 byte (including ending null character). While size of any type
of pointer is 2 byte (near pointer).
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