c operator question

7/27/2019 C Operator Question

1/6

C operator questions with answers

(1)

What will be output of the following program?

#includeintmain(){ floata=0.7; if(a


2/6

What will be output of the following program?#include

intmain(){ inti=5,j; j=++i+++i+++i; printf("%d %d",i,j); return0;}

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 finalvalue 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

---------------------------------------------------------


3/6

(5)


#include

voidmain(){ intx; x=10,20,30; printf("%d",x); return0;}

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 rightSince assignment operator (=) has more precedence than

comma operator .So = operator will be evaluated first

than comma operator. In the following expression

x = 10, 20, 30First 10 will be assigned to x then comma operator will

be evaluated.

-------------------------------------------------------

(9)


#include

intmain(){

intx=100,y=20,z=5; printf("%d %d %d");


4/6

return0;}

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 inprintf is data which is present in stack. So output is

revere order of declaration. So output will be 5 20 100.

---------------------------------------------------------

(11)


#include

intmain(){ inta; a=sizeof(!5.6); printf("%d",a); return0;}

Output:

Turbo C++ 3.0: 2

Turbo C ++4.5: 2

Linux GCC: 4Visual 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.

--------------------------------------------------------


5/6

(12)


#include

intmain(){ floata; (int)a= 45; printf("%d,a); return0;}

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)).

--------------------------------------------------------

(3)


#include

intmain(){ inti=1; i=2+2*i++; printf("%d",i); return0;}

Output:

Turbo C++ 3.0: 5

Turbo C ++4.5: 5


6/6

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

---------------------------------------------------------

c operator question

Documents