Question

2. Given the following declarations, explain why some of the following assignments are incorrect

int x = 10;

char y = 'A';

int *ptr1;

char *ptr2 = &y;

a. x = y;

b. ptr1 = ptr2;

c. ptr1 = &x;

d. x = &ptr1;

e. ptr2 = &x;

Thanks.

Answer 1

Answer::::

int x = 10;

char y = 'A';

int *ptr1;

char *ptr2 = &y;

These are given declarations...

For option a.) x = y;

If we assign x=y that is if we assign char to the integer variable if will print ASCII value of that character.

In the above example it will print ->> 65 which is ASCII value of 'A'.

For option b) ptr1 = ptr2;

Here ptr1 is integer pointer(int *ptr1) and ptr2 is character pointer (char *ptr2) so we cannot assign a character pointer to the integer pointer because it is assignment of incompatible types and if we print ptr1 then it will print garbage value.

for option c) ptr1 = &x;

Here we can assign a integer variable x (int x) to the integer pointer ptr1(int *ptr1) .The memory address of variable x is stored in pointer ptr1 and if we print ptr1 is shows the output as 10.

for option d) x = &ptr1;

Here we are assigning memory address of pointer ptr1 to the integer variable. Because of this assignment makes a integer from pointer (i.e memory location or address) without cast because we are storing it in a integer variable x.

If we print x then it will print garbage value.

For option e) ptr2 = &x;

Here we are storing memory address of integer variable x into pointer ptr2 . This assignment is of incompatible types.But if we print only ptr2 then it will print garbage value because incompatibility of their types.And if we print *ptr2 then it will print value at address of that variable in this it will print 10 which value of x.