CODEWITHSUNDEEP
c++
dast'
dast'

Reputation: 11

How do I read this small program?

I have been struggling to find out how increment operators relate to the value of b, c and the way reference sign changes the value when it is being written. The program prints out 8, but I am not sure what the reference actually does in this particular situation. Here is a code:

int fun1(int d) {
 ++d;
 return d++;
}
int fun2(int &d) {
 ++d;
 return d++;
}
int main(void) {
 int a = 1, b, c;
 b = fun1(a);
 c = fun2(b);
 cout << a + b + c << endl;
 return 0;
}

Thank you in advance!

Upvotes: 0

Views: 104

Answers (2)

TimB
TimB

Reputation: 1000

Since the first answer didn't fully go into the example at hand, I'll elaborate on this, because it's, at first, a tricky subject IMO.

int a = 1, b, c;

This declares three int variables: a, b and c. It also initializes a with the value 1. b and c receive the default start value for ints, so 0.

As the first answer mentioned, fun1(int d) copies the value of the variable passed to it into a local variable, but does not alter the original variable. What's the matter with ++d and d++, though? This got to do with the order in which operations get performed. ++d takes the value of d, increases it by one, then assigns it back to d. d++ on the other hand first assigns the value of d back to d (thus returning it from a function if used in a return statement) before increasing it by one. Why is this relevant?

Let's keep in mind fun1(a) works with a local variable so does not, in fact, alter variable a. ++d increases the local variable by 1, so it's now 2. Then d++ returns the current value of d (2), and then increases the local variable by 1 again -- but this doesnt't get stored anywhere outside of the function fun1(). Hence, the variable b has now the value 2.

c = fun2(b);

is a different case, though. Here, we pass a reference to the variable b into the function. So ++d doesn't just increase the value of a local variable but instead increases the value of b, it's now 3. d++ returns this value to c (so it gets assigned 3), but also increases b by 1 again to 4. Why? Because d is a reference to b.

Our ultimate values are hence: a = 1, b = 4, c = 3, and 1 + 4 + 3 = 8.

Upvotes: 0

Tarick Welling
Tarick Welling

Reputation: 3265

The crux of your problem lie in this part of your code:

int fun1(int d) { }
int fun2(int &d) { }

The int d in the first function are value semantics. This means that the value of the variable passed to function fun1 is copied and only know inside of fun1.

The int &d are reference semantics which means that the variable passed to fun2 is referenced by d and as such if you modify d inside of fun2 the actual value of the variable passed to fun2 is changed.

In your case this means that the value of a is copied by fun1 while fun2 updates the value of b. Because d is used in fun2 as a reference to the actual variable b wherein the value is stored.

And more to the point of the value of 8 printed by this program. ++d is the pre-increment operator, d++ is the post-increment operator. Meaning:

  • pre-increment: increment d first then use the new value of d
  • post-increment: keep the value of d and use that to do something and after that increment the value of d

Combining these things (value/reference semantics and the behaviour of pre/post increment) we get that that fun1 and fun2 return the original value of d which is incremented before the original value is returned. So return d++; is equivalent to this piece of code:

int returnvalue = d;
d = d + 1;
return returnvalue;

But because fun2 has a reference d to the variable b in main, the value in b is incremented by one because d was post-incremented in return d++.

ISOcpp value and reference semantics. And stackoverflow about the difference between pointers and references in C++.

If you need a much larger explanation of these terms and what they mean, this list of book recommendations is great. My choice would be A Tour of C++ by the creator of C++. It does a great job of going over everything you should now about C++ and is very cheap.

Upvotes: 4

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