Confusing outputs of references, how to analyse?

Question

So I don't really understand what references do. My homework assignment is to give the output for the following code. But A,B,H always change from being and int to int&.

a) A = int, B = int, H = int
b) A = int, B = int, H = int&
c) A = int, B = int&, H = int
d) A = int&, B = int, H = int
e) A = int&, B = int, H = int&
f) A = int&, B = int&, H = int

#include<iostream>

int foo (A a, B b) 
{
  H h = a; a = b;
  b = h; return h;
}

int main() 
{
  int a = 1;
  int b = 2;
  int r = foo (a, b);
  std::cout << a << " " << b << " " << r;

  return 0; 
}

So, a) was obviously not a problem. But already b) confused me.

for b); int& h = a is setting h's address to a's address? So h == Adresse of a, meaning value of h == a. Then a = b (where b == 2) which makes a == b == h == 2, so why is a == 1 in the output?

for c); h == 1 is clear, but the line a = b I find very confusing. b is a reference, not? Does that mean, that b's address is now the same as the address of a, meaning value of a == b? I'm guessing my thoughts are correct, because I get the proper output.

for d); h = reference of a so a has the same address like h? Then a has the name address as b? So a == b == 2. But now comes the weird part for me. b = h but b == 2, so 2 = h, but isn't that a conflict with L/R - values? And why is h now 1?

for e); no clue, int& h ?= int& a, what happens here? h and a have the same address? Then the address of a is now b? So the address is 2? Then b holds the address of h?

for f); First, h has now the address of a, so h == a. Then a has the same address as b. Then the address of b is the same as h. Correct thinking?

Solutions are

a) 1 2 1
b) 1 2 2
c) 1 1 1
d) 2 2 1
e) 2 2 2
f) 2 1 1

I need especially help with b) and e), thanks a lot and sorry for the long text.

Related: https://stackoverflow.com/questions/57483/what-are-the-differences-between-a-pointer-variable-and-a-reference-variable-in — Michael Chourdakis, Aug 01 '19 at 15:16
What type is `H`? Also `C` is not present in the snippet. `H` == `C`? — Felix Glas, Aug 01 '19 at 15:16
I think this is a good question underneath. The problem is I'm confused. There is no `C` in your code. — , Aug 01 '19 at 15:17

eerorika · Answer 1 · 2019-08-01T15:43:05.303

b)

for b); int& h = a is setting h's address to a's address?

The reference is not an object, and it does not have an address. h refers to a of foo.

So why is a == 1 in the output?

Because the parameter A a of foo is a separate object from int a of main function. H h refers to this local variable within foo and not the one in main. Assigning the local int within one function has no effect on a local variable of another function. a was initialised to 1 in main and that variable is not modified at all.

c)

b is a reference, not?

Yes. b within foo is a reference in case c).

Does that mean, that b's address is now the same as the address of a, meaning value of a == b?

Again, references are not objects and don't have addresses. b in foo refers to b in main.

d)

h = reference of a so a has the same address like h? Then a has the name address as b?

Again, references are not objects and don't have addresses. a in foo refers to a in main.

And why is h now 1?

Because that's what it is initialised to within foo.

e)

no clue, int& h ?= int& a, what happens here? h and a have the same address? Then the address of a is now b? So the address is 2? Then b holds the address of h?

Again, references are not objects and don't have addresses. The reference a in foo refers to the a in main given as an argument, and h refers to that same object. b is not a reference.

d)

First, h has now the address of a, so h == a. Then a has the same address as b. Then the address of b is the same as h.

h is not a reference. b in foo refers to the b in main, and a refers to a in main.

But I thought i.e. y = &x means y holds the address of x? Or does int& x mean something totally else?$ — Yves Betschmann, Aug 01 '19 at 15:56

Vlad from Moscow · Answer 2 · 2019-08-01T15:56:35.167

In the case b)

b) A = int, B = int, H = int&

neither original a nor b is changed in the function because the function deals with copies of the variables (the variables are passed by value).

h is of a referenced type that references the variable a inside the function.

 H h = a;

So in this statement

a = b;

the value of h will be the value of a after the assignment. So h will be equal to 2. And the output will be

b) 1 2 2

As for the case e)

e) A = int&, B = int, H = int&

then the only difference compared with the case b) is that the variable a is passed by reference. So not the only variable h but also the original variable a will be changed after this assignment

a = b;

So the output will be

e) 2 2 2

Confusing outputs of references, how to analyse?

2 Answers2