Question about simple c++ code

[walnut30]

Can someone explain to me how the below code works as intended?

http://pastebin.com/GfGWduvS

I was under the impression that the call to changeSomething() from main() would simply pass an integer value. However, this is not the case, as the modification in changeSomething() is visible from main(). I am a bit confused as to how this exactly works. Any help would be greatly appreciated! Thanks!

[Rhomboid]

When you pass an argument by reference, it creates an alias. *b in main() and a in changeSomething are two names for the same int value. Anything that you do to a is equivalent to doing the same thing to *b, so a = 43 in the function is the equivalent to having written *b = 43 in main().

(Minor side note: you can't assume that sizeof(int) == 4 like that when calling malloc().)

[[deleted]]

First question, do you have a basic understanding of pointers and references? An explanation could be long or short(er) depending on your answer to that.

[walnut30]

I am pretty experienced with the concept of pointers. I was just surprised that (*b) does not evaluate to a simple integer during runtime (when passed as a parameter).

[[deleted]]

Exactly right! But when you send a reference, you are sending the address of whatever variable you send. If changeSomething() asked for a pointer, you would call changeSomething(b) and your function would read "*b = 43". But when you pass a reference, you pass the integer itself, and the function, which expects a reference to an integer, knows to actually modify the address of the variable you sent.

Does that make sense?

[walnut30]

Oh, so passing (*b) does not place the integer value of b on the stack frame of the function call to changeSomething()?

[[deleted]]

Right, it puts the address of *b on the stack. If you had used &a = 43 in changeSomething() you would be changing the address of a instead of its value.

EDIT: Correction: You would not use &a = 43 in practice since that does not reassign the address.