Find the output of the following program.

void main()

{

printf(“sizeof (void *) = %d \n“, sizeof( void *));

            printf(“sizeof (int *)    = %d \n”, sizeof(int *));

            printf(“sizeof (double *)  = %d \n”, sizeof(double *));

            printf(“sizeof(struct unknown *) = %d \n”, sizeof(struct unknown *));

            }

Answer            :

sizeof (void *) = 2

sizeof (int *)    = 2

sizeof (double *)  =  2

sizeof(struct unknown *) =  2

Explanation:

The pointer to any type is of same size.

To get string input from the keyboard which one of the following is better?

           char inputString[100] = {0};

To get string input from the keyboard which one of the following is better?

            1) gets(inputString)

            2) fgets(inputString, sizeof(inputString), fp)

Answer & Explanation:

The second one is better because gets(inputString) doesn't know the size of the string passed and so, if a very big input (here, more than 100 chars) the charactes will be written past the input string. When fgets is used with stdin performs the same operation as gets but is safe.

Which version do you prefer of the following two,

1) printf(“%s”,str);      // or the more curt one

2) printf(str);

Answer & Explanation:

Prefer the first one. If the str contains any  format characters like %d then it will result in a subtle bug. 

Find the output of the following program.

void main()

{

int i=10, j=2;

int *ip= &i, *jp = &j;

int k = *ip/*jp;

printf(“%d”,k);

}         

Answer:

Compiler Error: “Unexpected end of file in comment started in line 5”.

Explanation:

The programmer intended to divide two integers, but by the “maximum munch” rule, the compiler treats the operator sequence / and * as /* which happens to be the starting of comment. To force what is intended by the programmer,

int k = *ip/ *jp;           

// give space explicity separating / and *

//or

int k = *ip/(*jp);

// put braces to force the intention 

will solve the problem.  

Find the output of the following program.

void main()

{

char ch;

for(ch=0;ch<=127;ch++)

printf(“%c   %d \n“, ch, ch);

}

Answer:

            Implementaion dependent

Explanation:

The char type may be signed or unsigned by default. If it is signed then ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is always smaller than 127.

Find the output of the following program.

Is this code legal?

int *ptr;

ptr = (int *) 0x400;

Answer:

                        Yes

Explanation:

The pointer ptr will point at the integer in the memory location 0x400.

Find the output of the following program.

         main()

{

char a[4]="HELLO";

printf("%s",a);

}         

Answer:

                        Compiler error: Too many initializers

Explanation:

The array a is of size 4 but the string constant requires 6 bytes to get stored.

Find the output of the program.

           main()

{         

char a[4]="HELL";

printf("%s",a);

}

Answer:

                        HELL%@!~@!@???@~~!

Explanation:

The character array has the memory just enough to hold the string “HELL” and doesnt have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it    accidentally comes across a NULL character.

Find the output of following program.

           main()

{

                        int a=10,*j;

            void *k;

                        j=k=&a;

            j++; 

                        k++;

            printf("\n %u %u ",j,k);

}

Answer:

                        Compiler error: Cannot increment a void pointer

Explanation:

Void pointers are generic pointers and they can be used only when the type is not known and as an intermediate address storage type. No pointer arithmetic can be done on it and you cannot apply indirection operator (*) on void pointers.

Find the output of following program.

main()

                        {

                                    extern int i;

                        {          int i=20;

                         {        

                           const volatile unsigned i=30; printf("%d",i);

                         }

                        printf("%d",i);

                        }

                          printf("%d",i);

            }         

            int i;

Printf can be implemented by using __________ list.

Answer:

                        Variable length argument lists

Find output of the following program.

           char *someFun()

            {

            char *temp = “string constant";

            return temp;

            }

            int main()

            {

            puts(someFun());

            }

Answer:

            string constant

Explanation:

            The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers. 

Find output of following program.

         
            char *someFun1()

            {

            char temp[ ] = “string";

            return temp;

            }

            char *someFun2()

            {

            char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};

            return temp;

            }

            int main()

            {

            puts(someFun1());

            puts(someFun2());

            }

Answer:

            Garbage values.

Explanation:

            Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.