Algorithms for problem solving - C programming MCQs with Answers

(1) Given in the following code segment:

	void main(void)
	{
		char x='\0';
		char n='N';
		printf("%u""%s\n",&n, &n);
	}
          

What will be the result of execution?

[A] ddddd N(where d represents any digit)
[B] 78 N
[C] 78 garbage
[D] compilation error

Comment

Answer: Option [A]

(2) Given in the following program:

	main()
	{
		int x=49;
		for (;x;)
		x--;
		printf("%d\n",x);
	}
	

What will be the output of the program?

[A] 49
[B] 0
[C] -49
[D] none of these

Comment

Answer: Option [B]

For all non-zero values of x the for will execute. x is decremented from 49 to 0. Hence 0 is printed in the screen.

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(3) Given in the following program:

	main()
	{
		int x=3, y=2;
		dp(x/y)
	}
          

What will be the output of the program?

[A] prints x/y=1
[B] prints #e=1.5
[C] prints #x/y=1
[D] none of tehes

Comment

Answer: Option [A]
(4) In the following code segment:

	int z,x=5,y=-10,a=4,b=2;
	z=x++ - --y* b/a;
	

What will be the final value of z?

[A] 5
[B] 6
[C] 10
[D] 11

Comment

Answer: Option [C]

The post increment operator is associated with the variable x. So the current variable is considered i.e. x=5.

The pre increment operator is associated with the variable y. So the value of y decreses by 1. That is y=-11.

Multiplication operator and division operator has the same precedence. So the first one is executed first.

So the expression is:

z=5-(-11)*2/4

=5-(-22/4)

=5-(-5)

=10

(5) Suppose that x is initialized as:

	short int x; /*assume x is 16 bits in size*/
	What is maximum number that can be printed using
	printf("%d\n",x);
	
[A] 127
[B] 128
[C] 255
[D] 32767

Comment

Answer: Option [D]

By default short integer is signed. 1 bit is reserveed for sign i.e. only 15 bits are available.

So, the largest number will be:

215+214+..........+upto 20

i.e. 32767

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