String questions in c interview 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).
30 comments:
how can we convert a number to octal format which are less than 8.
for eg.,what is the octal format of 6?
can any1 giv ans....
Hi...I'm a new member here. I need a solution to this simple Q. :Write a program that initializes a 10*3 array so that the first
element of each row contains a number, the second element contains
its square, and the third element contains its cube.
Start with 1 and stop at 10. For example, the first few rows will
look like this:
1,1,1
2,4,8
3,9,27
.
.
10, 100,1000
Next, prompt the user for a cube, look up this value in the table,
and report the cube’s root and the root’s square.
i have written a code for mithu_the reporter.
#include
void main()
{
int x[10][3]={0};
int f=1,n=1;
for(int i=0;i<10;i++)
{
for(int j=0;j<3;j++)
{
x[i][j]=f;
f=f*(i+1);
printf("%d ",x[i][j]);
}
n++;
f=n;
printf("\n");
}
}
write a program that compares two strings with out using strcmp().
plz explain 2nd prinf statement of Q:10,,,,,,,,,,,,,i hve understand rest of the printf statements.......a little bit confusion.........
Morvelous site that i never seen any where.Thanks to rithesh kumar for providing the most useful data.
i want to say from my heart a Big Bravo to the guys working on this code it is wonderful.
great job!!
hi ritesh u r website is too good,really great job man,can you explain the 10th question answer in detail,i have tried a lot but still i am not sure about the answer
Can you tell exactly which thing are you know getting.
printf("%d\t",***(ptr+1)); explain this
i am getting output as garbage value,
Numbers which are less than 8 will be remain same i.e. 6 in octal will remain 6.
can any1 tell to write a program that takes two strings as input passed as command line arguments within double quotes as shown below and outputs the concatenated string.
plzz rply fast
i have written a code for string merge sort.its dry run is correct ,but it does not give correc output.please correct this program and please mention where i have done wrong.
#include
#include
#include
void main()
{
clrscr();
char str[5][5],str1[6][6],temp[40][40];
int i,j,k;
printf("enter 1st array of string \n");
for(i=0;i<5;i++)
{
gets(str[i]);
}
printf("the first array of strings are=\n");
for(i=0;i<5;i++)
{
printf("\n str[%d]=",i);
printf("%s\t",str[i]);
}
printf("\n");
printf("enter 2nd array of string =\n");
for(j=0;j<6;j++)
{
gets(str1[j]);
}
for(j=0;j<6;j++)
{
printf("\n str1[%d]=",j);
printf("%s\t",str1[j]);
}
printf("\n");
i=0;
j=0;
k=0;
while((i<5)&&(j<6))
{
if(strcmp(str[i],str1[j])<0)
{
strcpy(temp[k],str[i]);
i++;
k++;
}
else
{
strcpy(temp[k],str1[j]);
j++;
k++;
}
}
while(i<5)
{
strcpy(temp[k],str[i]);
i++;
k++;
}
while(j<6)
{
strcpy(temp[k],str1[j]);
j++;
k++;
}
printf("\n");
printf("after sorting=\n");
printf("\n");
for(k=0;k<11;k++)
printf("%s\t",temp[k]);
getch();
}
can u give me code for program..
In java
If User enters numeric value then add or if he enters string then concatinate.
(ptr+1) point next two-dimensional array;
*(ptr+1) give base address of (&array+1) i.e next 2-d array(say array1);
**(ptr+1)-array1[0];
***(ptr+1)-array1[0][0];
which contains garbage value;
can u give me code for following example in c using functions
see the example is consider the string "ram_uma_raju_ram_ravi_uma" this the input sting and the program it will give the output as find the duplicate words along with count number i.e how many times that word repeated
output:"ram2_uma2_raju1_ravi1"
please give the code for this i need it urgent
c prog to search a char & report the position
how to sort a structure without using library functions
explain ques.15 answer
answer this
Write an interactive c program that will encode or decode a
line of text. To encode a line of text, proceed as follows:
Convert each character, including blank spaces, to its
ASCII equivalent.
Generate a positive random integer. Add this integer to the
ASCII equivalent of each character. The same random integer
will be used for the entire line of text.
Suppose that N1 represents the lowest permissible value in
the ASCII code, and N2 represents the highest permissible
value. If the number obtained in step 2 above exceeds N2,
then subtract the largest possible multiple of N2 from this
number, and add the remainder to N1. Hence the encoded
number will always fall between N1 and N2, and will
therefore always represent some ASCII character.
Display the characters that correspond to the encoded ASCII
values.
The procedure is reversed when decoding a line of text. Be
certain, however, that the same random number is used in
decoding as was used in encoding.
please explain all once
can you explain theses statements output
int a=5,k,b=5,t,c=5,r;
k=++a + ++a + a++;
printf("%i %i\n",a,k);
t=b++ + ++b + ++b;
printf("%i %i\n",b,t);
r=++c + ++c + ++c+ ;
printf("%i %i\n",c,r);
in gcc comipler
k=21
t=19
r=22
a=b=c=8
how can any body explain plzzzzzzzzz
k=++a + ++a + a++;
in this first of all
a++ => a=a+1;
a++ => a=6;
++a => a=6;
++a => a=7 for third
so value of a will be 7
and 7+7+7=21
C Programming Tips And Tricks...
Vist This Link..
https://youtu.be/TBKVuM9W8RM
write a program in c to reverse a words starting with r in a sentence
Q2: There will not be compilation error because bound checking for array is not done by compiler. It will print garbage value because the array stores 11 characters(The African) without null character
Value of k will be undefined...
On different complier value of k will be diff
The questions here are really conceptual based ,, and the explanation is also good
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