Sams Teach Yourself C++ in 21 Days

5. The indirection operator returns the value at the address stored in a pointer. The
   address-of operator (&) returns the memory address of the variable.


6. The const int ptrOnedeclares that ptrOneis a pointer to a constant integer.
   The integer itself cannot be changed using this pointer.
   The int const ptrTwodeclares that ptrTwois a constant pointer to integer.
   After it is initialized, this pointer cannot be reassigned.

Exercises ........................................................................................................

1. a. int pOne;declares a pointer to an integer.
   b. int vTwo;declares an integer variable.
   c. int pThree = &vTwo;declares a pointer to an integer and initializes it
   with the address of another variable,vTwo.


2. unsigned short pAge = &yourAge;
   3.pAge = 50;


3. The following is one possible answer:
   1: #include
   2:
   3: int main()
   4: {
   5: int theInteger;
   6: int pInteger = &theInteger;
   7: pInteger = 5;
   8:
   9: std::cout << “The Integer: “
   10: << *pInteger << std::endl;
   11:
   12: return 0;
   13: }
   5.pIntshould have been initialized. More importantly, because it was not initialized
   and was not assigned the address of any memory, it points to a random place in
   memory. Assigning a literal ( 9 ) to that random place is a dangerous bug.


4. Presumably, the programmer meant to assign 9 to the value at pVar, which would
   be an assignment to SomeVariable. Unfortunately, 9 was assigned to be the value
   of pVarbecause the indirection operator (*) was left off. This will lead to disaster
   if pVaris used to assign a value because it is pointing to whatever is at the address
   of 9 and not at SomeVariable.

834 Appendix D

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