C objective questions on coding to practice

These are the c++/cpp objective questions related to coding. these objective questions consists of programs to test your knowledge on syntax's. so learn the concepts and check these objective questions to test and improve your programming and error debugging skills. I am sharing 10+ objective questions related to programs in cpp in this post. practice all these objective questions and improve your programming skills.

C Objective Questions on Coding to Practice for beginners :-

1)

#include
main()
{
struct xx
{
int x=3;
char name[]=”hello”;
};
struct xx *s=malloc(sizeof(struct xx));
printf(“%d”,s->x);
printf(“%s”,s->name);
}

Answer: Compiler Error

Explanation:
Initialization should not be done for structure members inside the structure declaration

2)

main( )
{
printf(“%d”, out);
}
int out=100;

Answer:
Compiler error: undefined symbol out in function main.

Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

3)

main()
{
char string[]=”Hello World”;
display(string);
}
void display(char *string)
{
printf(“%s”,string);
}

Answer:
Compiler Error : Type mismatch in redeclaration of function display

Explanation :
In third line, when the function display is encountered, the compiler doesn’t know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

4)

#define int char
main()
{
int i=65;
printf(“sizeof(i)=%d”,sizeof(i));
}

Answer: sizeof(i)=1

Explanation:
Since the #define replaces the string int by the macro char

5)

main()
{
int i=10;
i=!i>14;
Printf (“i=%d”,i);
}

Answer: i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

6)

main()
{
printf(“\nab”);
printf(“\bsi”);
printf(“\rha”);
}

Answer: hai

Explanation:
\n – newline \b – backspace \r – linefeed(previous line)

7)

main()
{
int i;
printf(“%d”,scanf(“%d”,&i)); // value 10 is given as input here
}

Answer: 1

Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

8)

void main()
{
int const * p=5;
printf(“%d”,++(*p));
}

Answer: Compiler error: Cannot modify a constant value.

Explanation:
p is a pointer to a “constant integer”. But we tried to change the value of the “constant integer”.

(9)

What will be output if you will compile and execute the following c code?
#include
void main()
{
char *str;
scanf("%[^\n]",str);
printf("%s",str);
}
(a)It will accept a word as a string from user.
(b)It will accept a sentence as a string from user.
(c)It will accept a paragraph as a string from user.
(d)Compiler error

Ans: (b)

Explanation:
Task of % [^\t] is to take the stream of characters until it doesn’t receive new line character ‘\n’ i.e. enter button of your keyboard.

(10)

What will be output if you will compile and execute the following c code?
#include
void main()
{
int array[3]={5};
int i;
for(i=0;i<=2;i++)
printf("%d ",array[i]);
}
(a)5 garbage garbage (b)5 0 0
(c)5 null null (d)Compiler error

Ans: (b)

Explanation:
Storage class of an array which initializes the element of the array at the time of declaration is static. Default initial value of static integer is zero.

11)

What will be output if you will compile and execute the following c code? #include
void main( )
{
register int i,x;
scanf("%d",&i);
x=++i + ++i + ++i;
printf("%d",x);
}
(a)17 (b)18 (c)21 (d)Compiler error

Output: (d)

Explanation:
In c register variable stores in CPU it doesn’t store in RAM. So register variable have not any memory address. So it is illegal to write &a.

(12)

What will be output if you will compile and execute the following c code?
#include
void main(){
int a=-20;
int b=-3;
printf("%d",a%b);
}
(a)2 (b)-2 (c)18 (d)-18

Output: (b)

Explanation:
Sign of resultant of modular division depends upon only the sign of first operand.

13)

#include
int main()
{
goto abc;
printf("main");
}
void dispaly()
{
abc:
printf("display");
}
(a)main (b)display (c)maindisplay (d)Compiler error

Output: (d)

Explanation:
Label of goto cannot be in other function because control cannot move from one function to another function directly otherwise it will show compiler error: unreachable label

14)

#include
int main()
{
const int i=5;
i++;
printf("%d",i);
}
(a)5 (b)6 (c)0 (d)Compiler error

Output: (d)

Explanation:
We cannot modify the const variable by using increment operator.

(15)

#include
int main()
{
int array[2][2][3]={0,1,2,3,4,5,6,7,8,9,10,11};
printf("%d",array[1][0][2]);
}
(a)4 (b)5 (c)6 (d)7 (e)8

Output: e

Explanation:
array[1][0][2] means 1*(2*3)+0*(3)+3=9th element of array starting from zero i.e. 8.

16)

#include
void main()
{
int a=5,b=10,c=1;
if(a&&b>c)
printf("codershunt");
else
break;
}

output: error Misplaced break.

Explanation:
Keyword break is not syntactical part of if else statement. So we cannot use break keyword in if else statement. This keyword can be use in case of loop or switch case statement. Hence when you will compile above code compiler will show an error message: Misplaced break.

17)

#include
void main()
{
if(!printf("Mukesh Ambani"))
if(printf(" Lakashmi Mittal"))
}

Output: Mukesh Ambani

Explanation:
Return type of printf function is int. This function return a integral value which is equal to number of characters a printf function will print on console. First of all printf function will print Mukesh Ambani. Since it is printing 13 character so it will return 13. So,
!printf("Mukesh Ambani")
= !13= 0
In c language zero represents false. So if(0) is false so next statement which inside thebody of first if statement will not execute.

18)

#include
void main()
{
int x=1;
if(x)
printf("The Godfather");
x;
else
printf("%d",x);
}

Output: error: Misplace else

Explanation:
If you are not using { and } in if clause then you can write only one statement. Otherwiseit will cause of compilation error: Misplace else

19)

#include
void main()
{
if('\0');
else if(NULL)
printf("codershunt");
else;
}

output: no output

Explanation:
‘\0’ is null character constant. Its ASCII value is zero. if(0) means false so program control will check else if clause. NULL is macro constant which has been defined in stdio.h which also returns zero.

20)

main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}

Answer: H

Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so itsvalue is H. Again & references it to an address and * dereferences it to the value H.

Coders Hunt

C objective questions on coding to practice