This topic is not clear to me.

Question

Can someone please explain this code?
https://code.sololearn.com/cGWfWrWRPMe3/?ref=app

Answer 1

Basically y is another name for the variable x. If you know about pointers, this is a very similar thing. When y is declared &y=x, it is saying that the address of y is equal to x, meaning they are names for the same spot (address) in memory. Whenever you change y, you are changing the value at a certain space in memory, which x also happens to point at.

I hope this explanation makes sense. Happy coding :)

Answer 2

X becomes 5 right after the declaration of y.
int &y = x; y = 5;
So now x and y are variable names for the address where the value at that address is 5.
Then they become 7 due to the post-increment operator that is used twice.

Answer 3

The variable will still increment when it's 5 since the loop condition is less than OR EQUAL to 5. When x becomes 6, then the loop stops.

The & symbol might be a bit confusing here because it's used in the declaration. Try this instead:

int y, x;
y = &x;
// a pointer
int* pX;
pX = &x
// or like this
int* pY = y;

Answer 4

Thanks for the first part.X value is 0.How can x and y become 5? Which side of the equation is reference for this case?

Secodly, how does the x value become 7 after loop?

Answer 5

By the way, &y means the adress of y in the memory, not the value itself, isn't it? The dereference of the pointer, *p, gives the value, not the adress.

Answer 6

Loop operates for once, since x is already 5.by 6 , loop is over, no more increments?

Still confused about &y, it is the adress of variable, not the value?

I guess this & sign is something else than adress.

Answer 7

Y is increased by 1 in the loop for once and become 6.X value is updated in the loop and become 6.Then it is increased by one after loop.Is it correct?

If it is correct, it s interesting to see the x value is automatically updated.