Re: How to detect const reference to temporary issues at compile or runtime?

James Kanze <>
Wed, 1 Dec 2010 08:25:40 -0800 (PST)
On Dec 1, 2:18 am, Victor Bazarov <> wrote:

On 11/30/2010 8:56 PM, Clinton Mead wrote:

I've found recently that most of the errors in my C++ programs are of
a form like the following example:


class Z
  Z(int n) : n(n) {}
  int n;

class Y
  Y(const Z& z) : z(z) {}
  const Z& z;

class X
  X(const Y& y) : y(y) {}
  Y y;

class Big
    for (int i = 0; i< 1000; ++i) { a[i] = i + 1000; }
  int a[1000];

X get_x() { return X(Y(Z(123))); }

int main()
  X x = get_x();
  Big b;
  std::cout<< x.y.z.n<< std::endl;

OUTPUT: 1000

I would expect this program to output 123 (the value of x.y.z.n set in
get_x()) but the creation of "Big b" overwrites the temporary Z. As a
result, the reference to the temporary Z in the object Y is now
overwritten with Big b, and hence the output is not what I would

Simply put, your program has undefined behavior. You initialize the
data member of your 'Y' with a reference to a temporary. For a very
brief moment that reference is valid, then the temporary gets destroyed
and the reference becomes invalid. Any attempt to use it has undefined
behavior as the result.

I think he knows that. He's just wondering why compilers don't
warn in this simple case.

When I compiled this program with gcc 4.5 with the option "-Wall", it
gave no warning.

The compilers aren't *that* sophisticated. I don't know of any that
would exist that could determine the problem.

Using a reference parameter to initialize a reference member in
a constructor shouldn't be that hard to detect. The question is
whether it would result in too many false warnings: if your
"constract" says that the object passed in must live until the
end of the lifetime of the object being constructed, there's no
problem. Whether that's a frequent case or not in general,
I don't know. (It never occurs in my own code, because my
personal coding guidelines insist on using a pointer when an
argument must live beyond the end of the function: a pointer
means you need an lvalue, so trying to pass a temporary causes
a compiler error.)

James Kanze

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