有关功能和指针的功课需要帮助

Question

Asume that f() returns a reference to an integer ( int & f(); ), f()+=5; and f()=f()+5; are different, explain how and give pesudo code for f() to illustrate this difference.

If p is an int *p , what is the difference between, these two statement in C++:

if (p!=NULL && *p !=0).... 
if (*p !=0 && p !=NULL)....

Answer 1

In the first, you can have f() declare two static variables and a static pointer to one of them.

Then return them alternately for each call, something like (pseudo-code since it's homework):

def f():
    static var1 = 0;
    static var2 = 42;
    static pointer curr_var = reference of var1
    if curr_var == reference of var1:
        curr_var = reference of var2
    else:
        curr_var = reference of var1
    return curr_var

or, even worse:

def f():
    static var1 = array[1024];
    static idx = -1;
    idx = (idx + 1) % 100
    return reference of var1[idx]

For your second question, the hint is the difference between *p and p . For example, I wouldn't use the second one in the case where p itself could be NULL.

Answer 2

f()+=5 could be different from f() = f() +5 if f() returned a reference to a different integer at each call. This could only happen if f() read from some global variable which was different each time that it was called.

The difference between if (p!=NULL && *p !=0) and if (*p !=0 && *p !=NULL) is that the first one checks whether p is null and then checks whether the int that p points to is 0 . The second one only checks whether the int that p points to is 0 (and performs this check twice).