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Sample Technical Questions

C Sample Question

65. main()
    	{
    	int y;
    	scanf("%d",&y); // input given is 2000
    	if( (y%4==0 && y%100 != 0) || y%100 == 0 )
    	     printf("%d is a leap year");
    	else
    	     printf("%d is not a leap year");
    	}

    Answer:

    2000 is a leap year

    Explanation:

    An ordinary program to check if leap year or not.

66. #define max 5
    	#define int arr1[max]
    	main()
    	{
    	typedef char arr2[max];
    	arr1 list={0,1,2,3,4};
    	arr2 name="name";
    	printf("%d %s",list[0],name);
    	}

    Answer:

    Compiler error (in the line arr1 list = {0,1,2,3,4})

    Explanation:

    arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.

    Rule of Thumb:

    #defines are used for textual replacement whereas typedefs are used for declaring new types.

67. int i=10;
    	main()
    	{
     	 extern int i;
              	  {
    	     int i=20;
    		{
    		 const volatile unsigned i=30;
    		 printf("%d",i);
    		}
    	      printf("%d",i);
    	   }
    	printf("%d",i);
    	}

    Answer:

    30,20,10

    Explanation:

    '{' introduces new block and thus new scope. In the innermost block i is declared as,

    const volatile unsigned

    which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

68. main()
    	{
    	    int *j;
    	    {
    	     int i=10;
    	     j=&i;
    	     }
    	     printf("%d",*j);
    }

    Answer:

    10

    Explanation:

    The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

69. main()
    	{
    	int i=-1;
    	-i;
    	printf("i = %d, -i = %d \n",i,-i);
    	}

    Answer:

    i = -1, -i = 1

    Explanation:

    -i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

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