Data types interview questions and answers with explanation
Note: As you know size of data types is
compiler dependent in c. Answer of all question is based upon that
compilers whose word size is two byte. Just to know you, size of data type in 2 bytes compilers is as follow:
char : 1 byte
int : 2 byte
float : 4 byte
double : 8 byte
Please adjust the answers according to size of data types in you compiler.
Frequently asked technical objective types multiple choice data types questions of placement in c programming language 1.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
printf( "%d\t" , sizeof (6.5));
printf( "%d\t" , sizeof (90000));
printf( "%d" , sizeof ( 'A' )); return 0;
}
Choose all that apply:
(E)Explanation:
Turbo C++ 3.0: 8 4 2
Turbo C ++4.5: 8 4 2
Linux GCC: 8 4 4
Visual C++: 8 4 4
By default data type of numeric constants is:
6.5 : double
90000: long int
‘A’: char
In C size of data type varies from compiler to compiler.
In TURBO C 3.0 (16 bit compilers) size of:
double is 8 byte
Long int is 4 byte
Character constant is 2 byte (size of char data type is one byte)
In TURBO C 4.5 or Linux GCC compilers (32 bit compilers) size of:
double is 8 byte
long int is 8 byte
Character constant is 2 byte
2.
Consider on following declaring of enum.
(i) enum cricket {Gambhir,Smith,Sehwag}c;
(ii) enum cricket {Gambhir,Smith,Sehwag};
(iii) enum {Gambhir,Smith=-5,Sehwag}c;
(iv) enum c {Gambhir,Smith,Sehwag};
Choose correct one:
(E)Explanation:
Syntax of enum data type is:
enum [<tag_name>]{
<enum_constanat_name> [=<integer_ value>],
…
} [<var_name>,…]
Note:
[] : Represents optional .
<>: Represents any valid c identifier
3.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
signed x;
unsigned y;
x = 10 +- 10u + 10u +- 10;
y = x;
if (x==y)
printf( "%d %d" ,x,y);
else if (x!=y)
printf( "%u %u" ,x,y); return 0;
}
Choose all that apply: Explanation:
Turbo C++ 3.0: 0 0
Turbo C ++4.5: 0 0
Linux GCC: 0 0
Visual C++: 0 0
Consider on the expression:
x = 10 +- 10u + 10u +- 10;
10: It is signed integer constant.
10u: It is unsigned integer constant.
X: It is signed integer variable.
In any binary operation of dissimilar data type for example: a + b
Lower
data type operand always automatically type casted into the operand of
higher data type before performing the operation and result will be
higher data type.
As we know operators enjoy higher precedence than binary operators. So our expression is:
x = 10 + (-10u) + 10u + (-10);
= 10 + -10 + 10 + (-10);
= 0
Note: Signed is higher data type than unsigned int.
So, Corresponding signed value of unsigned 10u is +10
4.
Which of the following is not modifier of data type in c?
Explanation:
To know more about these go to following link:
5.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
double num=5.2;
int var=5;
printf( "%d\t" , sizeof (!num));
printf( "%d\t" , sizeof (var=15/2));
printf( "%d" ,var); return 0;
}
Choose all that apply:
(C)Explanation:
Turbo C++ 3.0: 2 2 5
Turbo C ++4.5: 2 2 5
Linux GCC: 4 4 5
Visual C++: 4 4 5
sizeof(Expr) operator always returns the an integer value which represents the size of the final value of the expression expr.
Consider on the following expression:
!num
=!5.2
=0
0 is int type integer constant and it size is 2 by in TURBO C 3.0 compiler and 4 in the TURBO C 4.5 and Linux GCC compilers.
Consider on the following expression:
var = 15/2
=> var = 7
=> 7
7 is int type integer constant.
Any
expression which is evaluated inside the sizeof operator its scope
always will be within the sizeof operator. So value of variable var will
remain 5 in the printf statement.
6.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
const int *p;
int a=10;
p=&a;
printf( "%d" ,*p); return 0;
}
Choose all that apply:
(B)Explanation:
Turbo C++ 3.0: 10
Turbo C ++4.5: 10
Linux GCC: 10
Visual C++: 10
In the following declaration
const int *p;
p can keep address of constant integer.
7.
Consider on following declaration:
(i) short i=10;
(ii) static i=10;
(iii) unsigned i=10;
(iv) const i=10;
Choose correct one: Explanation:
Default data type of above all declaration is int.
8.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
int a= sizeof ( signed ) + sizeof ( unsigned );
int b= sizeof ( const )+ sizeof ( volatile );
printf( "%d" ,a+++b); return 0;
}
Choose all that apply:
(C)Explanation:
Turbo C++ 3.0: 8
Turbo C ++4.5: 8
Linux GCC: 16
Visual C++: 16
Default data type of signed, unsigned, const and volatile is int. In turbo c 3.0 size of int is two byte.
So, a = 4 and b =4
Now, a+++b
= a++ + b
= 4 + 4 //due to post increment operator.
=8
Note: In turbo c 4.5 and Linux gcc compiler size of int is 4 byte so your out will be 16
9.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
signed x,a;
unsigned y,b;
a=( signed )10u;
b=( unsigned )-10;
y = ( signed )10u + ( unsigned )-10;
x = y;
printf( "%d %u\t" ,a,b);
if (x==y)
printf( "%d %d" ,x,y);
else if (x!=y)
printf( "%u %u" ,x,y); return 0;
}
Choose all that apply:
(D)Explanation:
Turbo C++ 3.0: 10 65526 0 0
Turbo C ++4.5: 10 65526 0 0
Linux GCC: 10 4294967286 0 0 Visual C++: 10 4294967286 0 0
a=( signed )10u;
signed value of 10u is +10
so, a=10
b=( unsigned )-10;
unsigned value of -10 is :
MAX_VALUE_OF_UNSIGNED_INT – 10 + 1
In turbo c 3.0 complier max value of unsigned int is 65535
So, b = 65526
y = ( signed )10u + ( unsigned )-10;
= 10 + 65526 = 65536 = 0 (Since 65536 is beyond the range of unsigned int. zero is its corresponding cyclic vlaue)
X = y = 0
10.
Which of the following is integral data type?
Explanation:
In c char is integral data type. It stores the ASCII value of any character constant.
11.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
volatile int a=11;
printf( "%d" ,a); return 0;
}
Choose all that apply: Explanation:
Turbo C++ 3.0: We cannot predict
Turbo C ++4.5: We cannot predict
Linux GCC: We cannot predict
Visual C++: We cannot predict
We cannot predict the value of volatile variable because its value can be changed by any microprocessor interrupt.
12.
What is the range of signed int data type in that compiler in which size of int is two byte?
(E)Explanation:
Note: Size of int is always equal to word length of micro preprocessor in which your compiler has based.
13.
What will be output when you will execute following c code?
#include <stdio.h>
const enum Alpha{
X ,
Y =5,
Z
}p=10;
int main(){
enum Alpha a,b;
a= X ;
b= Z ;
printf( "%d" ,a+b-p); return 0;
}
Choose all that apply:
(A)Explanation:
Turbo C++ 3.0: -4
Turbo C ++4.5: -4
Linux GCC: -4
Visual C++: -4
Default value of enum constant X is zero and
Z = Y + 1 = 5 + 1 = 6
So, a + b – p
=0 + 6 -10 = -4
14.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
char a=250;
int expr;
expr= a+ !a + ~a + ++a;
printf( "%d" ,expr); return 0;
}
Choose all that apply:
(D)Explanation:
Turbo C++ 3.0: -6
Turbo C ++4.5: -6
Linux GCC: -6
Visual C++: -6
char a = 250;
250 is beyond the range of signed char. Its corresponding cyclic value is: -6
So, a = -6
Consider on the expression:
expr= a+ !a + ~a + ++a;
Operator! , ~ and ++ have equal precedence. And it associative is right to left.
So, First ++ operator will perform the operation. So value a will -5
Now,
Expr = -5 + !-5 + ~-5 + -5
= -5 + !-5 + 4 - 5
= -5 + 0 + 4 -5
= -6
15.
Consider on order of modifiers in following declaration:
(i) char volatile register unsigned c;
(ii) volatile register unsigned char c;
(iii) register volatile unsigned char c;
(iv) unsigned char volatile register c;
Explanation:
Order of modifier of variable in c has not any significant.
Choose correct one:
16.
What will be output when you will execute following c code?
#include <stdio.h>
int main(){
int a=-5;
unsigned int b=-5u;
if (a==b)
printf( "Avatar" );
else
printf( "Alien" ); return 0;
}
Choose all that apply:
(A)Explanation:
Turbo C++ 3.0: Avatar
Turbo C ++4.5: Avatar
Linux GCC: Avatar
Visual C++: Avatar
int a=-5 ;
Here variable a is by default signed int.
unsigned int b=-5u;
Constant -5u will convert into unsigned int. Its corresponding unsigned int value will be :
65536 – 5 + 1= 65532
So, b = 65532
In any binary operation of dissimilar data type for example: a == b
Lower
data type operand always automatically type casted into the operand of
higher data type before performing the operation and result will be
higher data type.
In c signed int is higher data type than unsigned int. So variable b will automatically type casted into signed int.
So corresponding signed value of 65532 is -5
Hence, a==b
17.
What will be output when you will execute following c code?
#include <stdio.h>
extern enum cricket x;
int main(){
printf( "%d" ,x); return 0;
}
const enum cricket{
Taylor ,
Kallis =17,
Chanderpaul
}x= Taylor | Kallis & Chanderpaul ;
Choose all that apply:
(C)Explanation:
Turbo C++ 3.0: 16
Turbo C ++4.5: Compilation error
Linux GCC: Compilation error
Visual C++: 16
x=Taylor |Kallis &Chanderpaul
= 0 | 17 & 18
= 0 |(17 & 18)
//& operator enjoy higher precedence than |
=0 |16
=16
18.
Which of the following is not derived data type in c?
Explanation:
Enum is primitive data type.
19.
What will be output when you will execute following c code?
#include <stdio.h>
enum A{
x , y =5,
enum B{
p=10,q
}varp;
}varx;
int main(){
printf( "%d %d" ,x,varp.q); return 0;
}
Choose all that apply:
(E)Explanation: Turbo C++ 3.0: Compilation error
Turbo C ++4.5: Compilation error
Linux GCC: Compilation error
Visual C++: Compilation error
Nesting of enum constant is not possible in c.
20.
Consider on following declaration in c:
(i) short const register i=10;
(ii) static volatile const int i=10;
(iii) unsigned auto long register i=10;
(iv) signed extern float i=10.0;
Choose correct one:
(C)Explanation:
Option (III) is in correct due to we cannot specify two storage class auto and register in the declaration of any variable.
Option (iv)
is in correct due to we cannot use signed or unsigned modifiers with
float data type. In c float data type by default signed and it cannot be
unsigned.
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