Question

1. Explain the error.

char c = 'A';

double *p = &c;

2.

Give the value of the left-hand side variable in each assignment statement. Assume the lines are executed sequentially. Give brief explanation for each of your answer.

int main()

{

   char blocks[3] = {'A','B','C'};

   char *ptr = &blocks[0];

   char temp;

   temp = blocks[0];

   temp = *(blocks + 2);

   temp = *(ptr + 1);

   temp = *ptr;

   ptr = blocks + 1;

   temp = *ptr;

   temp = *(ptr + 1);

   ptr = blocks;

   temp = *++ptr;

   temp = ++*ptr;

   temp = *ptr++;

   temp = *ptr;

   return 0;

}

Answer 1

1)

c is declared as a charachter (char data type)

while declaring a pointer it is important for the compiler to know what data type it is pointing at

hence , data type of the pointer is always given as the data type of the variable which is to be pointed by the pointer

the correct code will be

char c = 'A';

char*p = &c;

2)

int main()

{

   char blocks[3] = {'A','B','C'};

   char *ptr = &blocks[0]; // address of first element of the array (address where A is stored)

// adding '&' at the start of any variable name will give the address of that variable

   char temp;

   temp = blocks[0]; // 'A' the value at zeroth position of the array is passed to temp

   temp = *(blocks + 2); // 'C'

// the name of the array can be used to get the address of the first element of the array

// *blocks will return the value at zeroth position of the array

// if a * is added at the start of the variable name and the variable is an address of some other variable

// the value of the other variable will be returned

   temp = *(ptr + 1); // 'B'

// ptr had the address of the first element of the array form the second line of main function

// adding 1 to the address of second element will give the adress of second element

.// * the address of second varable will return the second element of the array

   temp = *ptr; // address of first element of the array

   ptr = blocks + 1; // address of the second element

   temp = *ptr; // 'B'

// as ptr was given the address of the second element , *ptr will give the contents of the address in ptr

   temp = *(ptr + 1); // 'C'

// adding 1 to ptr which had the address of second element will give the address of third element

// adding a * to it will give the value at the address of third element ( which is third element itself)

   ptr = blocks; // address of first element

   temp = *++ptr; // 'B'

// ptr was given address of first element in previous line

// increment '++' at beginning means the increment the value first then execute rest of the line

// the value of ptr is incremented to next address i.e. address of second element

// * gives the value of second element

   temp = ++*ptr; // 'C'

// *ptr will give 'B' but incrementation of 1 will make it C

// not because next element in the array is C

// its is due to the fact that adding 1 to ascii code of B will give the ascii code of C

// the value of ptr will still remain the address of the second element

   temp = *ptr++; // B

// increment operater is after the variable name

// so the statement is executed first and then the incrementation takes place

// ptr had the address of the second element

// the value has updated to the address of third element after the execution of the statement

   temp = *ptr; // C

// the value of ptr was updated to the address of the third element of the array after the execution of previous statement

   return 0;

}