C quiz with solution

1) 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.

2)   #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.

3) 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.

4) 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.

5) 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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