output question

Code:

---

#include <iostream>

int& test(int& i) {
  ++i;
  return i;
}

int main() {
  int i = 0;
  std::cout << test(i) << ' ' << test(i) << '\n';
}

---

I'm having trouble understanding the output from this program. Can someone explain to me why the output is "2 2" if test returns by value and "2 1" when it returns by reference?

Quote:

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Originally Posted by jw232

Code:

---

#include <iostream>

int& test(int& i) {
  ++i;
  return i;
}

int main() {
  int i = 0;
  std::cout << test(i) << ' ' << test(i) << '\n';
}

---

I'm having trouble understanding the output from this program. Can someone explain to me why the output is "2 2" if test returns by value and "2 1" when it returns by reference?

---

What seems to be happening is you are passing i (which is not a pointer) to your function which is now interpreting it as a reference. What is passed back is actualy the _address_ of i, which is then altered by the second test(i).

This is simply a pointer mistake, you might want to brush up on them. i DOES at some point have the value 1, however it's adress is changed to 2. The behavior of cout << is not a stream, it sends a whole data buffer rather then data fragments, so what you're actualy saying here is:

Code:

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std::cout << address of i << ' ' << address of i << '\n';

---

Which is why the output is 2,2. If you want to correct this, all you have to do is:

Code:

---

int test(int i) {
  return ++i; //You can turn this into one statement.
}

int main() {
  int i = 0;
  std::cout << test(i) << ' ' << test(i) << '\n';
}

---

Again, this was an issue with pointers, which you didint need to use.

Since the output is not as described, there is nothing to explain.

Edit: I feel the need to mention that no pointers were harmed during the execution of the code, since no pointers exist during the execution of this code. (References != pointers.)

I tried a different compiler and I got "2 1" for both cases. I guess it was just my compiler.

Quote:

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Originally Posted by jw232

I tried a different compiler and I got "2 1" for both cases. I guess it was just my compiler.

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You ... what? Are we looking at the same code here? Starts at zero, increments inside the function? I'm pretty sure it's not possible to get 2 1 out of either case, let alone both. (Unless this is going to turn out to be one of those trick sequence-point things. Now I have to look up <<.)

Have you tried it yourself?
G++ in Cygwin gives me "2 1" for value, "2 2" for reference
VC++ gives me "2 1" both cases

I don't see what you mean, test is called twice, therefore i is incremented twice.

I'm using g++. I get "1 2" and "1 1".

OOH! This is fun!
Using the int & version:

Code:

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g++ -O3 -o temp.exe temp.cpp

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gives "1 2".

Code:

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g++ -o temp.exe temp.cpp

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gives "2 2". So it must be one of those sequence-point things involved in the << line. (I hate those things.)
And using Visual C++ (don't remember the optimization level) I did get "2 1".

Splitting the line into two statements (thus forcing the sequence point issue) I got the "1\n2" I expected.

Edit: I would have been less surprised by this had I fully thought through the return of int &, I think. For some reason, I was thinking "int test(int &i)". Oops.

Test is modifying main's version of i two times, and using the result. The code is basically equivalent to this:

Code:

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std::cout << ++i << ' ' << ++i << '\n';

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You are not allowed to modify a value twice in the same unbroken expression. Nor can a value be retrieved and modified in the same expression, unless it is retrieved to calculate it's new value.

Sequence points created by operators &&, ||, ?:, and ,(comma) break up an expression such that order across them is defined. Expressions can also be broken into separate lines, thereby being separate expressions entirely. << is not a sequencing operator, so it does not guarantee order, and thus the problem.

King Mir got this one off the bat. It's just a dressed up sequence point violation, and thus a bunch of undefined output.