C operator questions with answers


C programming tricky objective type operators questions and answers with explanation for written test and interview


(1)
What will be output of the following program?

#include<stdio.h>
int main(){
    float a=0.7;d 
    if(a<0.7){
         printf("C");
    }
    else{
         printf("C++");
    }
    return 0;
}


Explanation

Output: 

Turbo C++ 3.0: c

Turbo C ++4.5: c

Linux GCC: c

Visual C++: c


Explanation: 
0.7 is double constant (Default). Its binary value is written in 64 bit.

Binary value of 0.7 = (0.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 )

Now here variable a is a floating point variable while 0.7 is double constant. So variable a will contain only 32 bit value i.e.

a = 0.1011 0011 0011 0011 0011 0011 0011 0011 while
0.7 = 0.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011....
It is obvious a < 0.7


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(2)
What will be output of the following program?
        
#include<stdio.h>
int main(){
    int i=5,j;
    j=++i+++i+++i;
    printf("%d %d",i,j);
    return 0;
}

Explanation

Output: 

Turbo C++ 3.0: 8 24

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error


Explanation:

Rule :- ++ is pre increment operator so in any arithmetic expression it first increment the value of variable by one in whole expression then starts assigning the final value of variable in the expression.

Compiler will treat this expression j = ++i+++i+++i; as
i = ++i + ++i + ++i;

Initial value of i = 5 due to three pre increment operator final value of i=8.
Now final value of i i.e. 8 will assigned to each variable as shown in the following figure:


So, j=8+8+8
j=24 and
i=8


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(3)
What will be output of the following program?

#include<stdio.h>
int main(){
    int i=1;
    i=2+2*i++;
    printf("%d",i);
    return 0;
}

Explanation

Output: 

Turbo C++ 3.0: 5

Turbo C ++4.5: 5

Linux GCC: 5

Visual C++: 5


Explanation:
i++ i.e. when postfix increment operator is used any expression the it first assign the its value in the expression the it increments the value of variable by one. So,
i = 2 + 2 * 1
i = 4
Now i will be incremented by one so i = 4 + 1 = 5


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(4)
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);
    return 0;
}

Explanation

Output: 

Turbo C++ 3.0: 0

Turbo C ++4.5: 0

Linux GCC: 0

Visual C++: 0


Explanation:
== is relational operator which returns only two values.
0: If a == b is false
1: If a == b is true
Since
a=2
b=7
So, a == b is false hence b=0


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(5)
What will be output of the following program?

#include<stdio.h>
void main(){
    int x;
    x=10,20,30;
    printf("%d",x);
    return 0;
}

Explanation

Output: 

Turbo C++ 3.0: 10

Turbo C ++4.5: 10

Linux GCC: 10

Visual C++: 10


Explanation :
Precedence table:

Operator
Precedence
Associative
 =
More than ,
Right to left
 ,
Least
Left to right

Since assignment operator (=) has more precedence than comma operator .So = operator will be evaluated first than comma operator. In the following expression
x = 10, 20, 30
First 10 will be assigned to x then comma operator will be evaluated.


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(6)
What will be output of the following program?

#include<stdio.h>
int main(){
    int a=0,b=10;
    if(a=0){
         printf("true");
    }
    else{
         printf("false");
    }
    return 0;
}

Explanation

Output: 

Turbo C++ 3.0: false

Turbo C ++4.5: false

Linux GCC: false

Visual C++: false


Explanation:
As we know = is assignment operator not relation operator. So, a = 0 means zero will assigned to variable a. In c zero represent false and any non-zero number represents true.
So, if(0) means condition is always false hence else part will execute.


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(7)
What will be output of the following program?

#include<stdio.h>
int main(){
    int a;
    a=015 + 0x71 +5;
    printf("%d",a);
    return 0;
}

Explanation

Output: 

Turbo C++ 3.0: 131

Turbo C ++4.5: 131

Linux GCC: 131

Visual C++: 131


Explanation:
015 is octal number its decimal equivalent is = 5 * 8 ^ 0 + 1 * 8 ^ 1 = 5 + 8 = 13
0x71 is hexadecimal number (0x is symbol of hexadecimal) its decimal equivalent is = 1 * 16 ^ 0 + 7 * 16 ^ 1 = 1 + 112 = 113
So, a = 13 + 113 + 5 = 131


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(8)
What will be output of the following program?

#include<stdio.h>
int main(){
    printf("%d %d %d",sizeof(3.14),sizeof(3.14f),sizeof(3.14L));
    return 0;
}

Explanation

Output: 

Turbo C++ 3.0: 8 4 10

Turbo C ++4.5: 8 4 10

Linux GCC: 8 4 12

Visual C++: 8 4 8

Explanation:
3.14f is floating point constant. Its size is 4 byte. 3.14 is double constant (default). Its size is 8 byte. 3.14L is long double constant. Its size is 10 byte. sizeof() operator always return the size of data type which is written inside the(). It is keyword.



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(9)
What will be output of the following program?

#include<stdio.h>
int main(){
    int x=100,y=20,z=5;
    printf("%d %d %d");
    return 0;
}

Explanation

Output:

Turbo C++ 3.0: 5 20 100

Turbo C ++4.5: 5 20 100

Linux GCC: Garbage values

Visual C++: 5 100 20

By default x, y, z are auto type data which are stored in stack in memory. Stack is LIFO data structure. So in stack first stores 100 then 20 then 5 and program counter will point top stack i.e. 5. Default value of %d in printf is data which is present in stack. So output is revere order of declaration. So output will be 5 20 100.


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(10)
What will be output of the following program?

#include<stdio.h>        
int main(){
    int a=2;
    a=a++ + ~++a;
    printf("%d",a);
    return 0;
}

Explanation

Output: 

Turbo C++ 3.0: -1

Turbo C ++4.5: 0

Linux GCC: 0

Visual C++: 0


Explanation:
Same theory as question (2) and (13).


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(11)
What will be output of the following program?

#include<stdio.h>
int main(){
    int a;
    a=sizeof(!5.6);
    printf("%d",a);
    return 0;
}

Explanation

Output:
Turbo C++ 3.0: 2
Turbo C ++4.5: 2
Linux GCC: 4
Visual C++: 4

Explanation:
! is negation operator it return either integer 0 or 1.
! Any operand = 0 if operand is non zero.
! Any operand = 1 if operand is zero.
So, !5.6 = 0
Since 0 is integer number and size of integer data type is two byte.


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(12)
What will be output of the following program?

#include<stdio.h>
int main(){
    float a;
    (int)a= 45;
    printf("%d,a);
    return 0;
}

Explanation

Output:
Turbo C++ 3.0: Compilation error
Turbo C ++4.5: Compilation error
Linux GCC: Compilation error
Visual C++: Compilation error

Explanation:
After performing any operation on operand it always return some constant value.

(int) i.e. type casting operator is not exception for this. (int) a will return one constant value and we cannot assign any constant value to another constant value in c.

(int)a = 45; is equivalent to
3456 = 45 ( Here 3456 in any garbage value of int(a)).


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(13)
What will be output of the following program?

#include<stdio.h>
int main(){
     int i=5;
     int a=++i + ++i + ++i;
     printf("%d",a);
     return 0;
}

Explanation

Output: 

Turbo C++ 3.0: 21

Turbo C ++4.5: 21

Linux GCC: 22

Visual C++: 24


Explanation:
Rule : ++ (in ++i) is pre increment operator so in any arithmetic expression it first increment the value of variable by one in whole equation up to break point then start assigning the value of variable in the equation. There are many break point operators in. For example:

(1) Declaration statement.
(2) && or operator.
(3) Comma (,) operator etc.

In the following expression:
int a=++i + ++i + ++i;

Here break point is due to declaration .It break after each increment i.e. (initial value of i=5) after first increment value 6 assign to variable i then in next increment will occur and so on.
So, a = 6 + 7 + 8;


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Looping questions



If you have any queries or suggestions in above C operator question with answers, please suggest us.

67 comments:

Anonymous said...

The Questions was gud we require more tough Questions

Anonymous said...

please clearly explain me how the output
for question 10 is 0 or -1 .........

Anonymous said...

plz xplain 13 question

Rahul Soni said...

Hiiiii,
Ques no 2 & 13 show contradictory concepts.
a = 2;
d = ++a + ++a + ++a;
printf("d: %d",d);
O/P:Ques_02:- 15
O/P:Ques_13:- 12

why and how ?????
Plz clear this doubt........

Thanks:-)

Sagun gautam said...

can we add address of any variable we like inside the pointer. If so then we cud edit the value of any memory address we like.

sanjay kumar said...

when we write the c program as:
main()
{
clrscr();
int a=10;
printf("output%d",a);
}
here generate error in clrscr() function.please tell me deeply why generate error..

ritesh kumar said...

Hi Sanjiv,

In C language declaration statement must before the definition.

so correct way to write it:

main()
{
int a=10;
clrscr();

printf("output%d",a);
}

DATTATRAYA OZARKAR said...

can any one tell me logic for exchange of two number using only 2 variable

anilgontla said...

swaping without using third variable

int a,b;
a=a+b;
b=a-b;
a=a-b;
now a and b are swaped

Abul said...

Can anyone explain the difference between the question 2 & 13? I could not understand why the difference is there ?

Pl explain me

Suresh said...

In tubo c 3.0,The answer for question 13 is 24 only because the a value incremented by 3times before you run the program and the final value 8 will be stored in the variable a;so j=++i + (++i) + (++i)8+8+8=24.

Mukesh said...

in example 13, instead of writing
int a=++i+++i+++i, we declare a separately, we would get same answer as in question 2...i.e.
first declare int a;
then write the expression
a=++i+++i+++i;

Anonymous said...

a=a++ + ~++a; /* here a=2;
in the case of post increment 1st a++ not assign...and ++a comes to picture and final value of a=3;
that a=3 assign to a++ and ++a.
so
a=3 + ~3;
a=-1;
and than after increment of a++ so a=0;
ans is a=0.

Anonymous said...

I Like It......... GP

hafeehathim said...

Hi,
There are many break up points in C. Here the break up point is declaration..
see this example...
int a=2,d;
d=++a + ++a + ++a;
printf("%d",d);
Output: 5+5+5=15 (d is declared first and then value is assigned)

int a=2;
int d=++a + ++a + ++a;
printf("%d",d);
Output: 3+4+5=12 (d is declared and assigned in a single statement)

hafeehathim said...

Hi,
the answer for question 13 is 21. because the variable is declared and assigned in a single statement. here the break up point is declaration.

Anonymous said...

hey how it becomes a=3 + ~3= -1 it is 0

Ritu Agarwal said...

question no.9
What will be output of the following program?

#include
int main(){
int x=100,y=20,z=5;
printf("%d %d %d");
return 0;
}
ans
the given solution of this question is wrong
actually during the run of the program it'll give garbage value not correct ans

Anonymous said...

no no.. its is not zero.. ~3=-4 it returns the 2's cmpliment

Rohit Lanjekar said...

explain 2 and 13...i got confused

rakesh said...

Hi Ritu,
Actually the solution i.e output is compiler dependent.. i am using turbo C compiler ... and the solution is correct ... it prints whatever is recently present on the top of the stack...

Anonymous said...

int a=10;
int b=20;
a=a+b;
b=a-b;
a=a-b;
Printf("%d\n %d\n",a,b);

Anonymous said...

no no ....
a=2 + ~3 (in preincrement a will be used first then will be incremented )
a=2 + -4
a= -2
now we will increment a by 1 as I stated above
a= -2 +1
a=-1

Anonymous said...

20
10

Anonymous said...

Please explain again 2 and 13

Anonymous said...

Q. no.1
compilation error: d undeclared
line 3 : float a=0.7;d

Anonymous said...

Sir, please give some problem on Bitwise Operators..

muhammad masood said...

what does its means ~3 and << >> in c language

MAHENDRA KALSANGRA said...

how to find binary value of 32bit OR 64bit number..?
in above how to write .7 in 64 bit binary number...?

Anonymous said...

Arithmetically :
~a = -a -1

sophitorium engineering college said...

wap to overload +/-/>>/<< operator

Anonymous said...

please give more examples with explanation about bit wise opertors

Abdullah Naeem said...

Suresh ...
in code :: block 12.11(Application)
The answer for question 13 ... only because the a value incremented by 3or2 times before you run the program and the final value 8 will be stored in the variable a; but j=++i + (++i) + (++i)8+8+8=22 but not 24
that why tell me it?

Ahmed Iraqi said...

When I write this code in Turbo c++ 4.5 compiler will display 8 , but i cannot understand .. please help>>
code :
# include
void main()
{
int a=2;
a=++a + ++a;

printf("%d",a);
}

Manav said...

.7x2=1.4 (1)
.4x2=.8 (0)
.8x2=1.6 (1)
.6x2=1.2 (1)
.2x2=.4 (0)
.4x2=.8 (0)
.8x2=1.6 (1)
.6x2=1.2 (1)
.
.
.
so on

.7 binary will be = 0.1011 0011.......

emon said...

how can i find ceiling and floor without using library funtion ,u can use conditional operator

emon said...

#include
int max_(int x,int y);

int main ( )
{
int a,b,c,d;
scanf("%d%d%d%d", &a, &b, &c, &d);
int z=max_(a,b);
int m=max_(z,c);
int n=max_(m,d);
printf("max number is : %d",n);
return 0;
}

int max_(int x,int y)
{

int z=(x>y) ? x : y ;
return z;

}
pls fulfill this program,i cant get the output of ceiling .i want output ceiling without using library funtion

Kanav Singh Badwal said...

a=++a + ++a
a= 4 + 4
a= 8

Anonymous said...

int a=1,b=0,c;
c= b++ || a++ && --a;
printf("%d %d %d ",a,b,c);

plz help me in this program...

timesh kb said...

negative value of3 it mean -3

timesh kb said...

i has a break point so it increment imeediately
a=++a + ++a
a=3 + 4// there is break point so it inc immediately after 3 to 4
a=7

timesh kb said...

a=1
b=1
c=1

koushik ahamed said...

hafeehathim@thanks good point

eric948470 said...

For question 3, I got 4 as the output, compiling with gcc.

eric948470 said...

For question 10, I got -3 as the output, compiling with gcc, which is different from the given answer.

Tanga vinod said...

These are all called bitwise rightshift, leftshift and complement operators. if u want to perform operation on this if u want to convert to binary then only it is possible

Tanga vinod said...

These are all called bitwise rightshift, leftshift and complement operators. if u want to perform operation on this if u want to convert to binary then only it is possible

Kartheek Yenugula said...

For question 3, I got 4 as the output check it

Kartheek Yenugula said...

For question 3, I got 4 as the output check it

nimi said...

why answer of 2 and 13 different

tejaswi bandla said...

int t,a=10,b=20;

t=a;
a=b;
b=t;
print the values of a and b then also we get the values swapped
this is with help of third variable

ayush mittal said...

what will be value of i
i=1;
i=i+++++i;

asif mujawar said...


int a=20,b=30;
a=a^b; //a=20,b=30
after bitwise ex-or a wil become 10
b=a^b; //a=10,b=30
after bitwise ex-or b wil become 20
a=a^b; //a=10,b=20
after bitwise ex-or a wil become 30

Unnikrishnan.R said...

because its mandatory in c to declare varibles first
main()
{
int a=10;
clrscr();
printf("output%d",a);
}

Jitendra Kumar said...

Good Questions!
What will be the value of the given expression:
s = ++a - ++a + a-- - ++a where a = 1

Kix Mr said...

Tks very much for your post.

Avoid surprises — interviews need preparation. Some questions come up time and time again — usually about you, your experience and the job itself. We've gathered together the most common questions so you can get your preparation off to a flying start.

You also find all interview questions at link at the end of this post.

Source: Top 10 interview questions and answers

Best rgs

Unknown said...

plzzzz plzz can anyone give a detailed and easy solution of Q.2 and Q.13

bantu shishupal singh said...

in gcc int i=2;
i=i++ + ~++i;// 2 + -5(binary"100""sign 1's compliment is 1011 so -8+3=-5")
o/p -3

rajkamal said...

@ tejasvi
only two variables should be used not three variables.....

Madhuri Ravipati said...
This comment has been removed by the author.
ROHIT GUPTA said...

ansr 12 is rit one.....becoz pre increament opr fst increase the value of the varriable and the assign the value one bye one so thts why 2x is rit ansr............initially i=2 when first ++i occur thn it increase the value 3 and assign the value 3 in equation and after agggin it increase in sme scenario and assign the value 4 and then 5 thn after we get 12................

Shwetha Kamath said...

Linux answer for Q13
a= ++i + ++i + ++i
associativity of ++ operator is right to left, hence, a= ++i + (++i + ++i)

inside brackets, both operators are incremented before adding, hence totally, it is incremented twice and becomes 7
a=++i + (7+7)
a= ++i + 14

Now i is incremented again and becomes 8

a=8+14
a=22
:)

Shwetha Kamath said...
This comment has been removed by the author.
Shwetha Kamath said...

Can someone please explain how we get 0 for Question 10 in linux?

Mangesh Ghote said...

can any one please explain below code:

#include
int main(){
int a=2;
a=a++ + ~++a;
printf("%d",a);
return 0;
}

RICHA SINGH said...

In question number 2, as per the explanation given the value of j comes out to be 24,but when i am executing the same code,I am getting answer as 22,i.e. the value of j as 22.

(j=++i + ++i + ++i;)
Please explain

mehul said...

what is lvalue required ? can anyone tell it?